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Notely Maths 12
Class 12 Maths
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3.4 Q-1

Question Statement

Evaluate the following integrals by parts:

Background and Explanation

In these problems, we will use the integration by parts formula, which is: ∫u,dv=uv−∫v,du\int u , dv = uv - \int v , du

The goal is to break down complex integrals into simpler components by selecting appropriate functions for uu and dvdv.

For these problems, functions like sin⁡x\sin x, ln⁡x\ln x, and powers of xx will be paired with their derivatives to apply integration by parts efficiently.

Solution

i. ∫xsin⁡x,dx\int x \sin x , dx

We start by choosing u=xu = x and dv=sin⁡x,dxdv = \sin x , dx.

u=x,du=dx,dv=sin⁡x,dx,v=−cos⁡x.\begin{aligned} u = x, & \quad du = dx, dv = \sin x , dx, & \quad v = -\cos x. \end{aligned}

Applying the formula:

∫xsin⁡x,dx=x(−cos⁡x)−∫(−cos⁡x),dx=−xcos⁡x+∫cos⁡x,dx=−xcos⁡x+sin⁡x+C.\int x \sin x , dx = x (-\cos x) - \int (-\cos x) , dx = -x \cos x + \int \cos x , dx = -x \cos x + \sin x + C.

Thus, the solution is: sin⁡x−xcos⁡x+C.\sin x - x \cos x + C.

ii. ∫ln⁡x,dx\int \ln x , dx

We choose u=ln⁡xu = \ln x and dv=1,dxdv = 1 , dx.

u=ln⁡x,du=1x,dx,dv=1,dx,v=x.\begin{aligned} u = \ln x, & \quad du = \frac{1}{x} , dx, dv = 1 , dx, & \quad v = x. \end{aligned}

Applying the formula:

∫ln⁡x,dx=xln⁡x−∫x⋅1x,dx=xln⁡x−∫1,dx=xln⁡x−x+C.\int \ln x , dx = x \ln x - \int x \cdot \frac{1}{x} , dx = x \ln x - \int 1 , dx = x \ln x - x + C.

Thus, the solution is: xln⁡x−x+C.x \ln x - x + C.

iii. ∫xln⁡x,dx\int x \ln x , dx

We choose u=ln⁡xu = \ln x and dv=x,dxdv = x , dx.

u=ln⁡x,du=1x,dx,dv=x,dx,v=x22.\begin{aligned} u = \ln x, & \quad du = \frac{1}{x} , dx, dv = x , dx, & \quad v = \frac{x^2}{2}. \end{aligned}

Applying the formula:

∫xln⁡x,dx=x22ln⁡x−∫x22⋅1x,dx=x22ln⁡x−12∫x,dx.\int x \ln x , dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} , dx = \frac{x^2}{2} \ln x - \frac{1}{2} \int x , dx.

Now, integrating the remaining term:

x22ln⁡x−12⋅x22+C=x22ln⁡x−x24+C.\frac{x^2}{2} \ln x - \frac{1}{2} \cdot \frac{x^2}{2} + C = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C.

Thus, the solution is: x22(ln⁡x−12)+C.\frac{x^2}{2} \left( \ln x - \frac{1}{2} \right) + C.

iv. ∫x2ln⁡x,dx\int x^2 \ln x , dx

We choose u=ln⁡xu = \ln x and dv=x2,dxdv = x^2 , dx.

u=ln⁡x,du=1x,dx,dv=x2,dx,v=x33.\begin{aligned} u = \ln x, & \quad du = \frac{1}{x} , dx, dv = x^2 , dx, & \quad v = \frac{x^3}{3}. \end{aligned}

Applying the formula:

∫x2ln⁡x,dx=x33ln⁡x−∫x33⋅1x,dx=x33ln⁡x−13∫x2,dx.\int x^2 \ln x , dx = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} , dx = \frac{x^3}{3} \ln x - \frac{1}{3} \int x^2 , dx.

Now, integrating the remaining term:

x33ln⁡x−13⋅x33+C=x33ln⁡x−x39+C.\frac{x^3}{3} \ln x - \frac{1}{3} \cdot \frac{x^3}{3} + C = \frac{x^3}{3} \ln x - \frac{x^3}{9} + C.

Thus, the solution is: x33(ln⁡x−13)+C.\frac{x^3}{3} \left( \ln x - \frac{1}{3} \right) + C.

v. ∫x3ln⁡x,dx\int x^3 \ln x , dx

We choose u=ln⁡xu = \ln x and dv=x3,dxdv = x^3 , dx.

u=ln⁡x,du=1x,dx,dv=x3,dx,v=x44.\begin{aligned} u = \ln x, & \quad du = \frac{1}{x} , dx, dv = x^3 , dx, & \quad v = \frac{x^4}{4}. \end{aligned}

Applying the formula:

∫x3ln⁡x,dx=x44ln⁡x−∫x44⋅1x,dx=x44ln⁡x−14∫x3,dx.\int x^3 \ln x , dx = \frac{x^4}{4} \ln x - \int \frac{x^4}{4} \cdot \frac{1}{x} , dx = \frac{x^4}{4} \ln x - \frac{1}{4} \int x^3 , dx.

Now, integrating the remaining term:

x44ln⁡x−14⋅x44+C=x44ln⁡x−x416+C.\frac{x^4}{4} \ln x - \frac{1}{4} \cdot \frac{x^4}{4} + C = \frac{x^4}{4} \ln x - \frac{x^4}{16} + C.

Thus, the solution is: x44(ln⁡x−14)+C.\frac{x^4}{4} \left( \ln x - \frac{1}{4} \right) + C.

vi. ∫x4ln⁡x,dx\int x^{4} \ln x , dx

We use integration by parts, selecting u=ln⁡xu = \ln x and dv=x4dxdv = x^4 dx. The derivatives and integrals are:

u=ln⁡x,du=1xdx,dv=x4dx,v=x55.\begin{aligned} u = \ln x, & \quad du = \frac{1}{x} dx, dv = x^4 dx, & \quad v = \frac{x^5}{5}. \end{aligned}

Applying the integration by parts formula:

∫x4ln⁡x,dx=x55ln⁡x−∫x55⋅1x,dx\int x^4 \ln x , dx = \frac{x^5}{5} \ln x - \int \frac{x^5}{5} \cdot \frac{1}{x} , dx

Simplifying the second term:

=x55ln⁡x−15∫x4,dx= \frac{x^5}{5} \ln x - \frac{1}{5} \int x^4 , dx

Now, integrate x4x^4:

15∫x4,dx=x525\frac{1}{5} \int x^4 , dx = \frac{x^5}{25}

Thus, the solution becomes:

x55ln⁡x−x525+C=x55(ln⁡x−15)+C.\frac{x^5}{5} \ln x - \frac{x^5}{25} + C = \frac{x^5}{5} \left( \ln x - \frac{1}{5} \right) + C.

vii. ∫tan⁡−1x,dx\int \tan^{-1} x , dx

We apply integration by parts, selecting u=tan⁡−1xu = \tan^{-1} x and dv=dxdv = dx. The derivatives and integrals are:

u=tan⁡−1x,du=1x2+1,dx,dv=dx,v=x.\begin{aligned} u = \tan^{-1} x, & \quad du = \frac{1}{x^2 + 1} , dx, dv = dx, & \quad v = x. \end{aligned}

Applying the formula:

∫tan⁡−1x,dx=xtan⁡−1x−∫xx2+1,dx\int \tan^{-1} x , dx = x \tan^{-1} x - \int \frac{x}{x^2 + 1} , dx

To solve the second integral, notice that:

∫xx2+1,dx=12ln⁡(x2+1)\int \frac{x}{x^2 + 1} , dx = \frac{1}{2} \ln(x^2 + 1)

Thus, the solution is:

xtan⁡−1x−12ln⁡(x2+1)+C.x \tan^{-1} x - \frac{1}{2} \ln(x^2 + 1) + C.

viii. ∫x2sin⁡x,dx\int x^2 \sin x , dx

We use integration by parts twice. First, choose u=x2u = x^2 and dv=sin⁡x,dxdv = \sin x , dx. The derivatives and integrals are:

u=x2,du=2x,dx,dv=sin⁡x,dx,v=−cos⁡x.\begin{aligned} u = x^2, & \quad du = 2x , dx, dv = \sin x , dx, & \quad v = -\cos x. \end{aligned}

Applying the formula:

∫x2sin⁡x,dx=−x2cos⁡x+∫2xcos⁡x,dx.\int x^2 \sin x , dx = -x^2 \cos x + \int 2x \cos x , dx.

Now, integrate 2xcos⁡x2x \cos x by parts, choosing u=xu = x and dv=cos⁡x,dxdv = \cos x , dx. The derivatives and integrals are:

u=x,du=dx,dv=cos⁡x,dx,v=sin⁡x.\begin{aligned} u = x, & \quad du = dx, dv = \cos x , dx, & \quad v = \sin x. \end{aligned}

Applying the formula again:

∫2xcos⁡x,dx=2(xsin⁡x−∫sin⁡x,dx)=2(xsin⁡x+cos⁡x).\int 2x \cos x , dx = 2 \left( x \sin x - \int \sin x , dx \right) = 2 \left( x \sin x + \cos x \right).

Thus, the full solution is:

−x2cos⁡x+2(xsin⁡x+cos⁡x)+C=−x2cos⁡x+2xsin⁡x+2cos⁡x+C.-x^2 \cos x + 2(x \sin x + \cos x) + C = -x^2 \cos x + 2x \sin x + 2 \cos x + C.

ix. ∫x2tan⁡−1x,dx\int x^2 \tan^{-1} x , dx

We apply integration by parts, selecting u=tan⁡−1xu = \tan^{-1} x and dv=x2,dxdv = x^2 , dx. The derivatives and integrals are:

u=tan⁡−1x,du=1x2+1,dx,dv=x2,dx,v=x33.\begin{aligned} u = \tan^{-1} x, & \quad du = \frac{1}{x^2 + 1} , dx, dv = x^2 , dx, & \quad v = \frac{x^3}{3}. \end{aligned}

Applying the formula:

∫x2tan⁡−1x,dx=x33tan⁡−1x−∫x33⋅1x2+1,dx.\int x^2 \tan^{-1} x , dx = \frac{x^3}{3} \tan^{-1} x - \int \frac{x^3}{3} \cdot \frac{1}{x^2 + 1} , dx.

We divide x3x^3 by x2+1x^2 + 1:

x3x2+1=x−xx2+1.\frac{x^3}{x^2 + 1} = x - \frac{x}{x^2 + 1}.

Thus, the integral becomes:

∫x2tan⁡−1x,dx=x33tan⁡−1x−13∫x,dx+13∫xx2+1,dx.\int x^2 \tan^{-1} x , dx = \frac{x^3}{3} \tan^{-1} x - \frac{1}{3} \int x , dx + \frac{1}{3} \int \frac{x}{x^2 + 1} , dx.

We know the integral of xx is x22\frac{x^2}{2}, and ∫xx2+1,dx=12ln⁡(x2+1)\int \frac{x}{x^2 + 1} , dx = \frac{1}{2} \ln(x^2 + 1), so we get:

∫x2tan⁡−1x,dx=x33tan⁡−1x−x26+16ln⁡(x2+1)+C.\int x^2 \tan^{-1} x , dx = \frac{x^3}{3} \tan^{-1} x - \frac{x^2}{6} + \frac{1}{6} \ln(x^2 + 1) + C.

x. ∫xtan⁡−1x,dx\int x \tan^{-1} x , dx

We apply integration by parts, selecting u=tan⁡−1xu = \tan^{-1} x and dv=x,dxdv = x , dx. The derivatives and integrals are:

u=tan⁡−1x,du=1x2+1,dx,dv=x,dx,v=x22.\begin{aligned} u = \tan^{-1} x, & \quad du = \frac{1}{x^2 + 1} , dx, dv = x , dx, & \quad v = \frac{x^2}{2}. \end{aligned}

Applying the formula:

∫xtan⁡−1x,dx=x22tan⁡−1x−∫x22⋅1x2+1,dx.\int x \tan^{-1} x , dx = \frac{x^2}{2} \tan^{-1} x - \int \frac{x^2}{2} \cdot \frac{1}{x^2 + 1} , dx.

We simplify the second integral:

x2x2+1=1−1x2+1.\frac{x^2}{x^2 + 1} = 1 - \frac{1}{x^2 + 1}.

Thus, the integral becomes:

∫xtan⁡−1x,dx=x22tan⁡−1x−12∫dx+12∫1x2+1,dx.\int x \tan^{-1} x , dx = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int dx + \frac{1}{2} \int \frac{1}{x^2 + 1} , dx.

We know the integrals:

∫dx=x,∫1x2+1,dx=tan⁡−1x.\int dx = x, \quad \int \frac{1}{x^2 + 1} , dx = \tan^{-1} x.

Thus, the final solution is:

x22tan⁡−1x−x2+12tan⁡−1x+C=12tan⁡−1x(x2+1)−x2+C.\frac{x^2}{2} \tan^{-1} x - \frac{x}{2} + \frac{1}{2} \tan^{-1} x + C = \frac{1}{2} \tan^{-1} x \left( x^2 + 1 \right) - \frac{x}{2} + C.

xi. ∫x3tan⁡−1x,dx\int x^{3} \tan^{-1} x , dx

We use integration by parts, taking u=tan⁡−1xu = \tan^{-1} x and dv=x3,dxdv = x^3 , dx. The derivatives and integrals are:

u=tan⁡−1x,du=1x2+1,dx,dv=x3,dx,v=x44.\begin{aligned} u &= \tan^{-1} x, & du &= \frac{1}{x^2 + 1} , dx, dv &= x^3 , dx, & v &= \frac{x^4}{4}. \end{aligned}

Applying the formula:

∫x3tan⁡−1x,dx=x44tan⁡−1x−∫x44⋅1x2+1,dx.\int x^3 \tan^{-1} x , dx = \frac{x^4}{4} \tan^{-1} x - \int \frac{x^4}{4} \cdot \frac{1}{x^2 + 1} , dx.

We split x4x2+1\frac{x^4}{x^2 + 1} into two parts:

x4x2+1=x2−1x2+1.\frac{x^4}{x^2 + 1} = x^2 - \frac{1}{x^2 + 1}.

Thus, the integral becomes:

∫x3tan⁡−1x,dx=x44tan⁡−1x−14∫x2,dx+14∫1x2+1,dx.\int x^3 \tan^{-1} x , dx = \frac{x^4}{4} \tan^{-1} x - \frac{1}{4} \int x^2 , dx + \frac{1}{4} \int \frac{1}{x^2 + 1} , dx.

Now we evaluate the remaining integrals:

∫x2,dx=x33,∫1x2+1,dx=tan⁡−1x.\int x^2 , dx = \frac{x^3}{3}, \quad \int \frac{1}{x^2 + 1} , dx = \tan^{-1} x.

Thus, the final result is:

∫x3tan⁡−1x,dx=x44tan⁡−1x−x312+14x−14tan⁡−1x+C.\int x^3 \tan^{-1} x , dx = \frac{x^4}{4} \tan^{-1} x - \frac{x^3}{12} + \frac{1}{4} x - \frac{1}{4} \tan^{-1} x + C.

xii. ∫x3cos⁡x,dx\int x^3 \cos x , dx

We apply integration by parts, selecting u=x3u = x^3 and dv=cos⁡x,dxdv = \cos x , dx. The derivatives and integrals are:

u=x3,du=3x2,dx,dv=cos⁡x,dx,v=sin⁡x.\begin{aligned} u &= x^3, & du &= 3x^2 , dx, dv &= \cos x , dx, & v &= \sin x. \end{aligned}

Using the formula:

∫x3cos⁡x,dx=x3sin⁡x−∫3x2sin⁡x,dx.\int x^3 \cos x , dx = x^3 \sin x - \int 3x^2 \sin x , dx.

Now, we perform integration by parts again on ∫x2sin⁡x,dx\int x^2 \sin x , dx, taking u=x2u = x^2 and dv=sin⁡x,dxdv = \sin x , dx. The derivatives and integrals are:

u=x2,du=2x,dx,dv=sin⁡x,dx,v=−cos⁡x.\begin{aligned} u &= x^2, & du &= 2x , dx, dv &= \sin x , dx, & v &= -\cos x. \end{aligned}

Thus, we have:

∫x2sin⁡x,dx=−x2cos⁡x+2∫xcos⁡x,dx.\int x^2 \sin x , dx = -x^2 \cos x + 2 \int x \cos x , dx.

Now, we integrate ∫xcos⁡x,dx\int x \cos x , dx by parts, with u=xu = x and dv=cos⁡x,dxdv = \cos x , dx. The derivatives and integrals are:

u=x,du=dx,dv=cos⁡x,dx,v=sin⁡x.\begin{aligned} u &= x, & du &= dx, dv &= \cos x , dx, & v &= \sin x. \end{aligned}

Thus:

∫xcos⁡x,dx=xsin⁡x−∫sin⁡x,dx=xsin⁡x+cos⁡x.\int x \cos x , dx = x \sin x - \int \sin x , dx = x \sin x + \cos x.

Substituting back:

∫x3cos⁡x,dx=x3sin⁡x−3(−x2cos⁡x+2(xsin⁡x+cos⁡x)).\int x^3 \cos x , dx = x^3 \sin x - 3 \left( -x^2 \cos x + 2(x \sin x + \cos x) \right).

Simplifying:

∫x3cos⁡x,dx=x3sin⁡x+3x2cos⁡x−6xsin⁡x−6cos⁡x+C.\int x^3 \cos x , dx = x^3 \sin x + 3x^2 \cos x - 6x \sin x - 6 \cos x + C.

xiii. ∫sin⁡−1x,dx\int \sin^{-1} x , dx

We apply integration by parts, taking u=sin⁡−1xu = \sin^{-1} x and dv=dxdv = dx. The derivatives and integrals are:

u=sin⁡−1x,du=11−x2,dx,dv=dx,v=x.\begin{aligned} u &= \sin^{-1} x, & du &= \frac{1}{\sqrt{1 - x^2}} , dx, dv &= dx, & v &= x. \end{aligned}

Using the formula:

∫sin⁡−1x,dx=xsin⁡−1x−∫x1−x2,dx.\int \sin^{-1} x , dx = x \sin^{-1} x - \int \frac{x}{\sqrt{1 - x^2}} , dx.

The remaining integral is straightforward:

∫x1−x2,dx=−1−x2.\int \frac{x}{\sqrt{1 - x^2}} , dx = -\sqrt{1 - x^2}.

Thus, the final result is:

∫sin⁡−1x,dx=xsin⁡−1x+1−x2+C.\int \sin^{-1} x , dx = x \sin^{-1} x + \sqrt{1 - x^2} + C.

xiv. ∫xsin⁡−1x,dx\int x \sin^{-1} x , dx

We use integration by parts, taking u=sin⁡−1xu = \sin^{-1} x and dv=x,dxdv = x , dx. The derivatives and integrals are:

u=sin⁡−1x,du=11−x2,dx,dv=x,dx,v=x22.\begin{aligned} u &= \sin^{-1} x, & du &= \frac{1}{\sqrt{1 - x^2}} , dx, dv &= x , dx, & v &= \frac{x^2}{2}. \end{aligned}

Using the formula:

∫xsin⁡−1x,dx=x22sin⁡−1x−∫x22⋅11−x2,dx.\int x \sin^{-1} x , dx = \frac{x^2}{2} \sin^{-1} x - \int \frac{x^2}{2} \cdot \frac{1}{\sqrt{1 - x^2}} , dx.

We split the remaining integral:

∫x21−x2,dx=∫1−(1−x2)1−x2,dx=∫11−x2,dx−∫11−x2,dx.\int \frac{x^2}{\sqrt{1 - x^2}} , dx = \int \frac{1 - (1 - x^2)}{\sqrt{1 - x^2}} , dx = \int \frac{1}{\sqrt{1 - x^2}} , dx - \int \frac{1}{\sqrt{1 - x^2}} , dx.

Thus, the remaining part simplifies to:

x22sin⁡−1x−(12∫1−x21−x2,dx).\frac{x^2}{2} \sin^{-1} x - \left( \frac{1}{2} \int \frac{1 - x^2}{\sqrt{1 - x^2}} , dx \right).

The resulting final answer is:

∫xsin⁡−1x,dx=x22sin⁡−1x−14sin⁡−1x+141−x2+C.\int x \sin^{-1} x , dx = \frac{x^2}{2} \sin^{-1} x - \frac{1}{4} \sin^{-1} x + \frac{1}{4} \sqrt{1 - x^2} + C.

xv. ∫exsin⁡xcos⁡x,dx\int e^x \sin x \cos x , dx

We apply the product-to-sum formula: sin⁡xcos⁡x=12sin⁡(2x)\sin x \cos x = \frac{1}{2} \sin(2x). Thus, we rewrite the integral as:

∫exsin⁡xcos⁡x,dx=12∫exsin⁡(2x),dx.\int e^x \sin x \cos x , dx = \frac{1}{2} \int e^x \sin(2x) , dx.

We use integration by parts, taking u=exu = e^x and dv=sin⁡(2x),dxdv = \sin(2x) , dx. The derivatives and integrals are:

u=ex,du=ex,dx,dv=sin⁡(2x),dx,v=−12cos⁡(2x).\begin{aligned} u &= e^x, & du &= e^x , dx, dv &= \sin(2x) , dx, & v &= -\frac{1}{2} \cos(2x). \end{aligned}

Using the formula:

∫exsin⁡(2x),dx=−12excos⁡(2x)+12∫excos⁡(2x),dx.\int e^x \sin(2x) , dx = -\frac{1}{2} e^x \cos(2x) + \frac{1}{2} \int e^x \cos(2x) , dx.

Now, we integrate ∫excos⁡(2x),dx\int e^x \cos(2x) , dx by parts, with u=exu = e^x and dv=cos⁡(2x),dxdv = \cos(2x) , dx:

u=ex,du=ex,dx,dv=cos⁡(2x),dx,v=12sin⁡(2x).\begin{aligned} u &= e^x, & du &= e^x , dx, dv &= \cos(2x) , dx, & v &= \frac{1}{2} \sin(2x). \end{aligned}

Thus:

∫excos⁡(2x),dx=12exsin⁡(2x)−12∫exsin⁡(2x),dx.\int e^x \cos(2x) , dx = \frac{1}{2} e^x \sin(2x) - \frac{1}{2} \int e^x \sin(2x) , dx.

Substitute back:

∫exsin⁡(2x),dx=−12excos⁡(2x)+12(12exsin⁡(2x)−12∫exsin⁡(2x),dx).\int e^x \sin(2x) , dx = -\frac{1}{2} e^x \cos(2x) + \frac{1}{2} \left( \frac{1}{2} e^x \sin(2x) - \frac{1}{2} \int e^x \sin(2x) , dx \right).

Thus, the final result is:

∫exsin⁡xcos⁡x,dx=12ex(−cos⁡2x+sin⁡2x)+C.\int e^x \sin x \cos x , dx = \frac{1}{2} e^x \left( -\cos 2x + \sin 2x \right) + C.

xi. ∫x3tan⁡−1x,dx\int x^{3} \tan^{-1} x , dx

We use integration by parts, taking u=tan⁡−1xu = \tan^{-1} x and dv=x3,dxdv = x^3 , dx. The derivatives and integrals are:

u=tan⁡−1x,du=1x2+1,dx,dv=x3,dx,v=x44.\begin{aligned} u &= \tan^{-1} x, & du &= \frac{1}{x^2 + 1} , dx, dv &= x^3 , dx, & v &= \frac{x^4}{4}. \end{aligned}

Applying the formula:

∫x3tan⁡−1x,dx=x44tan⁡−1x−∫x44⋅1x2+1,dx.\int x^3 \tan^{-1} x , dx = \frac{x^4}{4} \tan^{-1} x - \int \frac{x^4}{4} \cdot \frac{1}{x^2 + 1} , dx.

We split x4x2+1\frac{x^4}{x^2 + 1} into two parts:

x4x2+1=x2−1x2+1.\frac{x^4}{x^2 + 1} = x^2 - \frac{1}{x^2 + 1}.

Thus, the integral becomes:

∫x3tan⁡−1x,dx=x44tan⁡−1x−14∫x2,dx+14∫1x2+1,dx.\int x^3 \tan^{-1} x , dx = \frac{x^4}{4} \tan^{-1} x - \frac{1}{4} \int x^2 , dx + \frac{1}{4} \int \frac{1}{x^2 + 1} , dx.

Now we evaluate the remaining integrals:

∫x2,dx=x33,∫1x2+1,dx=tan⁡−1x.\int x^2 , dx = \frac{x^3}{3}, \quad \int \frac{1}{x^2 + 1} , dx = \tan^{-1} x.

Thus, the final result is:

∫x3tan⁡−1x,dx=x44tan⁡−1x−x312+14x−14tan⁡−1x+C.\int x^3 \tan^{-1} x , dx = \frac{x^4}{4} \tan^{-1} x - \frac{x^3}{12} + \frac{1}{4} x - \frac{1}{4} \tan^{-1} x + C.

xii. ∫x3cos⁡x,dx\int x^3 \cos x , dx

We apply integration by parts, selecting u=x3u = x^3 and dv=cos⁡x,dxdv = \cos x , dx. The derivatives and integrals are:

u=x3,du=3x2,dx,dv=cos⁡x,dx,v=sin⁡x.\begin{aligned} u &= x^3, & du &= 3x^2 , dx, dv &= \cos x , dx, & v &= \sin x. \end{aligned}

Using the formula:

∫x3cos⁡x,dx=x3sin⁡x−∫3x2sin⁡x,dx.\int x^3 \cos x , dx = x^3 \sin x - \int 3x^2 \sin x , dx.

Now, we perform integration by parts again on ∫x2sin⁡x,dx\int x^2 \sin x , dx, taking u=x2u = x^2 and dv=sin⁡x,dxdv = \sin x , dx. The derivatives and integrals are:

u=x2,du=2x,dx,dv=sin⁡x,dx,v=−cos⁡x.\begin{aligned} u &= x^2, & du &= 2x , dx, dv &= \sin x , dx, & v &= -\cos x. \end{aligned}

Thus, we have:

∫x2sin⁡x,dx=−x2cos⁡x+2∫xcos⁡x,dx.\int x^2 \sin x , dx = -x^2 \cos x + 2 \int x \cos x , dx.

Now, we integrate ∫xcos⁡x,dx\int x \cos x , dx by parts, with u=xu = x and dv=cos⁡x,dxdv = \cos x , dx. The derivatives and integrals are:

u=x,du=dx,dv=cos⁡x,dx,v=sin⁡x.\begin{aligned} u &= x, & du &= dx, dv &= \cos x , dx, & v &= \sin x. \end{aligned}

Thus:

∫xcos⁡x,dx=xsin⁡x−∫sin⁡x,dx=xsin⁡x+cos⁡x.\int x \cos x , dx = x \sin x - \int \sin x , dx = x \sin x + \cos x.

Substituting back:

∫x3cos⁡x,dx=x3sin⁡x−3(−x2cos⁡x+2(xsin⁡x+cos⁡x)).\int x^3 \cos x , dx = x^3 \sin x - 3 \left( -x^2 \cos x + 2(x \sin x + \cos x) \right).

Simplifying:

∫x3cos⁡x,dx=x3sin⁡x+3x2cos⁡x−6xsin⁡x−6cos⁡x+C.\int x^3 \cos x , dx = x^3 \sin x + 3x^2 \cos x - 6x \sin x - 6 \cos x + C.

xiii. ∫sin⁡−1x,dx\int \sin^{-1} x , dx

We apply integration by parts, taking u=sin⁡−1xu = \sin^{-1} x and dv=dxdv = dx. The derivatives and integrals are:

u=sin⁡−1x,du=11−x2,dx,dv=dx,v=x.\begin{aligned} u &= \sin^{-1} x, & du &= \frac{1}{\sqrt{1 - x^2}} , dx, dv &= dx, & v &= x. \end{aligned}

Using the formula:

∫sin⁡−1x,dx=xsin⁡−1x−∫x1−x2,dx.\int \sin^{-1} x , dx = x \sin^{-1} x - \int \frac{x}{\sqrt{1 - x^2}} , dx.

The remaining integral is straightforward:

∫x1−x2,dx=−1−x2.\int \frac{x}{\sqrt{1 - x^2}} , dx = -\sqrt{1 - x^2}.

Thus, the final result is:

∫sin⁡−1x,dx=xsin⁡−1x+1−x2+C.\int \sin^{-1} x , dx = x \sin^{-1} x + \sqrt{1 - x^2} + C.

xiv. ∫xsin⁡−1x,dx\int x \sin^{-1} x , dx

We use integration by parts, taking u=sin⁡−1xu = \sin^{-1} x and dv=x,dxdv = x , dx. The derivatives and integrals are:

u=sin⁡−1x,du=11−x2,dx,dv=x,dx,v=x22.\begin{aligned} u &= \sin^{-1} x, & du &= \frac{1}{\sqrt{1 - x^2}} , dx, dv &= x , dx, & v &= \frac{x^2}{2}. \end{aligned}

Using the formula:

∫xsin⁡−1x,dx=x22sin⁡−1x−∫x22⋅11−x2,dx.\int x \sin^{-1} x , dx = \frac{x^2}{2} \sin^{-1} x - \int \frac{x^2}{2} \cdot \frac{1}{\sqrt{1 - x^2}} , dx.

We split the remaining integral:

∫x21−x2,dx=∫1−(1−x2)1−x2,dx=∫11−x2,dx−∫11−x2,dx.\int \frac{x^2}{\sqrt{1 - x^2}} , dx = \int \frac{1 - (1 - x^2)}{\sqrt{1 - x^2}} , dx = \int \frac{1}{\sqrt{1 - x^2}} , dx - \int \frac{1}{\sqrt{1 - x^2}} , dx.

Thus, the remaining part simplifies to:

x22sin⁡−1x−(12∫1−x21−x2,dx).\frac{x^2}{2} \sin^{-1} x - \left( \frac{1}{2} \int \frac{1 - x^2}{\sqrt{1 - x^2}} , dx \right).

The resulting final answer is:

∫xsin⁡−1x,dx=x22sin⁡−1x−14sin⁡−1x+141−x2+C.\int x \sin^{-1} x , dx = \frac{x^2}{2} \sin^{-1} x - \frac{1}{4} \sin^{-1} x + \frac{1}{4} \sqrt{1 - x^2} + C.

xv. ∫exsin⁡xcos⁡x,dx\int e^x \sin x \cos x , dx

We apply the product-to-sum formula: sin⁡xcos⁡x=12sin⁡(2x)\sin x \cos x = \frac{1}{2} \sin(2x). Thus, we rewrite the integral as:

∫exsin⁡xcos⁡x,dx=12∫exsin⁡(2x),dx.\int e^x \sin x \cos x , dx = \frac{1}{2} \int e^x \sin(2x) , dx.

We use integration by parts, taking u=exu = e^x and dv=sin⁡(2x),dxdv = \sin(2x) , dx. The derivatives and integrals are:

u=ex,du=ex,dx,dv=sin⁡(2x),dx,v=−12cos⁡(2x).\begin{aligned} u &= e^x, & du &= e^x , dx, dv &= \sin(2x) , dx, & v &= -\frac{1}{2} \cos(2x). \end{aligned}

Using the formula:

∫exsin⁡(2x),dx=−12excos⁡(2x)+12∫excos⁡(2x),dx.\int e^x \sin(2x) , dx = -\frac{1}{2} e^x \cos(2x) + \frac{1}{2} \int e^x \cos(2x) , dx.

Now, we integrate ∫excos⁡(2x),dx\int e^x \cos(2x) , dx by parts, with u=exu = e^x and dv=cos⁡(2x),dxdv = \cos(2x) , dx:

u=ex,du=ex,dx,dv=cos⁡(2x),dx,v=12sin⁡(2x).\begin{aligned} u &= e^x, & du &= e^x , dx, dv &= \cos(2x) , dx, & v &= \frac{1}{2} \sin(2x). \end{aligned}

Thus:

∫excos⁡(2x),dx=12exsin⁡(2x)−12∫exsin⁡(2x),dx.\int e^x \cos(2x) , dx = \frac{1}{2} e^x \sin(2x) - \frac{1}{2} \int e^x \sin(2x) , dx.

Substitute back:

∫exsin⁡(2x),dx=−12excos⁡(2x)+12(12exsin⁡(2x)−12∫exsin⁡(2x),dx).\int e^x \sin(2x) , dx = -\frac{1}{2} e^x \cos(2x) + \frac{1}{2} \left( \frac{1}{2} e^x \sin(2x) - \frac{1}{2} \int e^x \sin(2x) , dx \right).

Thus, the final result is:

∫exsin⁡xcos⁡x,dx=12ex(−cos⁡2x+sin⁡2x)+C.\int e^x \sin x \cos x , dx = \frac{1}{2} e^x \left( -\cos 2x + \sin 2x \right) + C.

xvi. ∫xsin⁡xcos⁡x,dx\int x \sin x \cos x , dx

We start by using the double-angle identity for sine: sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2 \sin x \cos x. This simplifies the integral:

∫xsin⁡xcos⁡x,dx=12∫xsin⁡2x,dx\int x \sin x \cos x , dx = \frac{1}{2} \int x \sin 2x , dx

Now, we apply integration by parts, selecting u=xu = x and dv=sin⁡2x,dxdv = \sin 2x , dx. The derivatives and integrals are:

u=x,du=dx,dv=sin⁡2x,dx,v=−12cos⁡2x.\begin{aligned} u = x, & \quad du = dx, dv = \sin 2x , dx, & \quad v = -\frac{1}{2} \cos 2x. \end{aligned}

Using the integration by parts formula:

∫u,dv=uv−∫v,du,\int u , dv = uv - \int v , du,

we get:

12[−xcos⁡2x2+12∫cos⁡2x,dx].\frac{1}{2} \left[ -\frac{x \cos 2x}{2} + \frac{1}{2} \int \cos 2x , dx \right].

The integral of cos⁡2x\cos 2x is sin⁡2x2\frac{\sin 2x}{2}, so we arrive at:

∫xsin⁡xcos⁡x,dx=14[−xcos⁡2x+sin⁡2x]+C.\int x \sin x \cos x , dx = \frac{1}{4} \left[ -x \cos 2x + \sin 2x \right] + C.

Thus, the final result is:

∫xsin⁡xcos⁡x,dx=14(−xcos⁡2x+sin⁡2x)+C.\boxed{\int x \sin x \cos x , dx = \frac{1}{4} \left( -x \cos 2x + \sin 2x \right) + C.}

xvii. ∫xcos⁡2x,dx\int x \cos^2 x , dx

To solve this, we use the identity cos⁡2x=1+cos⁡2x2\cos^2 x = \frac{1 + \cos 2x}{2}, which simplifies the integral:

∫xcos⁡2x,dx=12∫x(1+cos⁡2x),dx=12(∫x,dx+∫xcos⁡2x,dx).\int x \cos^2 x , dx = \frac{1}{2} \int x (1 + \cos 2x) , dx = \frac{1}{2} \left( \int x , dx + \int x \cos 2x , dx \right).

We now handle each part separately.

First part: ∫x,dx\int x , dx

This is straightforward:

∫x,dx=x22.\int x , dx = \frac{x^2}{2}.

Second part: ∫xcos⁡2x,dx\int x \cos 2x , dx

We apply integration by parts, selecting u=xu = x and dv=cos⁡2x,dxdv = \cos 2x , dx. The derivatives and integrals are:

u=x,du=dx,dv=cos⁡2x,dx,v=12sin⁡2x.\begin{aligned} u = x, & \quad du = dx, dv = \cos 2x , dx, & \quad v = \frac{1}{2} \sin 2x. \end{aligned}

Using the integration by parts formula:

∫xcos⁡2x,dx=xsin⁡2x2−∫sin⁡2x2,dx=xsin⁡2x2+cos⁡2x4.\int x \cos 2x , dx = \frac{x \sin 2x}{2} - \int \frac{\sin 2x}{2} , dx = \frac{x \sin 2x}{2} + \frac{\cos 2x}{4}.

Thus, the complete solution is:

∫xcos⁡2x,dx=12(x22+xsin⁡2x2+cos⁡2x4)+C.\int x \cos^2 x , dx = \frac{1}{2} \left( \frac{x^2}{2} + \frac{x \sin 2x}{2} + \frac{\cos 2x}{4} \right) + C.

The final result is:

∫xcos⁡2x,dx=x24+xsin⁡2x4+cos⁡2x8+C.\boxed{\int x \cos^2 x , dx = \frac{x^2}{4} + \frac{x \sin 2x}{4} + \frac{\cos 2x}{8} + C.}

xviii. ∫xsin⁡2x,dx\int x \sin^2 x , dx

We use the identity sin⁡2x=1−cos⁡2x2\sin^2 x = \frac{1 - \cos 2x}{2}, which simplifies the integral:

∫xsin⁡2x,dx=12∫x(1−cos⁡2x),dx=12(∫x,dx−∫xcos⁡2x,dx).\int x \sin^2 x , dx = \frac{1}{2} \int x (1 - \cos 2x) , dx = \frac{1}{2} \left( \int x , dx - \int x \cos 2x , dx \right).

We now handle each part separately.

First part: ∫x,dx\int x , dx

This is straightforward:

∫x,dx=x22.\int x , dx = \frac{x^2}{2}.

Second part: ∫xcos⁡2x,dx\int x \cos 2x , dx

We already computed this in the previous part:

∫xcos⁡2x,dx=xsin⁡2x2+cos⁡2x4.\int x \cos 2x , dx = \frac{x \sin 2x}{2} + \frac{\cos 2x}{4}.

Thus, the complete solution is:

∫xsin⁡2x,dx=12(x22−xsin⁡2x2−cos⁡2x4)+C.\int x \sin^2 x , dx = \frac{1}{2} \left( \frac{x^2}{2} - \frac{x \sin 2x}{2} - \frac{\cos 2x}{4} \right) + C.

The final result is:

∫xsin⁡2x,dx=x24−xsin⁡2x4−cos⁡2x8+C.\boxed{\int x \sin^2 x , dx = \frac{x^2}{4} - \frac{x \sin 2x}{4} - \frac{\cos 2x}{8} + C.}

xix. ∫(ln⁡x)2,dx\int (\ln x)^2 , dx

We apply integration by parts, selecting u=(ln⁡x)2u = (\ln x)^2 and dv=dxdv = dx. The derivatives and integrals are:

u=(ln⁡x)2,du=2ln⁡xx,dx,dv=dx,v=x.\begin{aligned} u = (\ln x)^2, & \quad du = \frac{2 \ln x}{x} , dx, dv = dx, & \quad v = x. \end{aligned}

Using the integration by parts formula:

∫u,dv=uv−∫v,du,\int u , dv = uv - \int v , du,

we get:

x(ln⁡x)2−∫x⋅2ln⁡xx,dx=x(ln⁡x)2−2∫ln⁡x,dx.x (\ln x)^2 - \int x \cdot \frac{2 \ln x}{x} , dx = x (\ln x)^2 - 2 \int \ln x , dx.

The integral of ln⁡x\ln x is xln⁡x−xx \ln x - x, so:

∫(ln⁡x)2,dx=x(ln⁡x)2−2(xln⁡x−x).\int (\ln x)^2 , dx = x (\ln x)^2 - 2 \left( x \ln x - x \right).

Thus, the final result is:

∫(ln⁡x)2,dx=x(ln⁡x)2−2xln⁡x+2x+C.\boxed{\int (\ln x)^2 , dx = x (\ln x)^2 - 2x \ln x + 2x + C.}

xx. ∫ln⁡(tan⁡x)sec⁡2x,dx\int \ln (\tan x) \sec^2 x , dx

We use integration by parts, selecting u=ln⁡(tan⁡x)u = \ln (\tan x) and dv=sec⁡2x,dxdv = \sec^2 x , dx. The derivatives and integrals are:

u=ln⁡(tan⁡x),du=1tan⁡xsec⁡2x,dx,dv=sec⁡2x,dx,v=tan⁡x.\begin{aligned} u = \ln (\tan x), & \quad du = \frac{1}{\tan x} \sec^2 x , dx, dv = \sec^2 x , dx, & \quad v = \tan x. \end{aligned}

Using the integration by parts formula:

∫u,dv=uv−∫v,du,\int u , dv = uv - \int v , du,

we get:

ln⁡(tan⁡x)⋅tan⁡x−∫tan⁡x⋅1tan⁡xsec⁡2x,dx=tan⁡xln⁡(tan⁡x)−∫sec⁡2x,dx.\ln (\tan x) \cdot \tan x - \int \tan x \cdot \frac{1}{\tan x} \sec^2 x , dx = \tan x \ln (\tan x) - \int \sec^2 x , dx.

The integral of sec⁡2x\sec^2 x is tan⁡x\tan x, so:

∫ln⁡(tan⁡x)sec⁡2x,dx=tan⁡xln⁡(tan⁡x)−tan⁡x+C.\int \ln (\tan x) \sec^2 x , dx = \tan x \ln (\tan x) - \tan x + C.

Thus, the final result is:

∫ln⁡(tan⁡x)sec⁡2x,dx=tan⁡xln⁡(tan⁡x)−tan⁡x+C.\boxed{\int \ln (\tan x) \sec^2 x , dx = \tan x \ln (\tan x) - \tan x + C.}

xxi. ∫xsin⁡x1−x2,dx\int \frac{x \sin x}{\sqrt{1 - x^2}} , dx

To solve this, we use the substitution u=sin⁡−1xu = \sin^{-1} x, which simplifies the expression. The integral becomes:

∫xsin⁡x1−x2,dx=−∫(x1−x2)sin⁡−1x,dx.\int \frac{x \sin x}{\sqrt{1 - x^2}} , dx = -\int \left( \frac{x}{\sqrt{1 - x^2}} \right) \sin^{-1} x , dx.

Now, apply integration by parts, selecting u=sin⁡−1xu = \sin^{-1} x and dv=x1−x2,dxdv = \frac{x}{\sqrt{1 - x^2}} , dx. The derivatives and integrals are:

u=sin⁡−1x,du=11−x2,dx,dv=x1−x2,dx,v=1−x2.\begin{aligned} u = \sin^{-1} x, & \quad du = \frac{1}{\sqrt{1 - x^2}} , dx, dv = \frac{x}{\sqrt{1 - x^2}} , dx, & \quad v = \sqrt{1 - x^2}. \end{aligned}

Using the integration by parts formula:

∫u,dv=uv−∫v,du,\int u , dv = uv - \int v , du,

we get:

−(sin⁡−1x⋅1−x2−∫1−x2,dx)=−sin⁡−1x⋅1−x2+∫dx.- \left( \sin^{-1} x \cdot \sqrt{1 - x^2} - \int \sqrt{1 - x^2} , dx \right) = -\sin^{-1} x \cdot \sqrt{1 - x^2} + \int dx.

Therefore:

∫xsin⁡x1−x2,dx=−sin⁡−1x⋅1−x2+x+C.{\int \frac{x \sin x}{\sqrt{1 - x^2}} , dx = -\sin^{-1} x \cdot \sqrt{1 - x^2} + x + C.}

Key Formulas or Methods Used

  • Integration by Parts: ∫u,dv=uv−∫v,du\int u , dv = uv - \int v , du

  • This formula is applied iteratively in these problems to handle integrals involving logarithms, trigonometric functions, and polynomials.

Summary of Steps

  1. Identify the parts of the integral to apply integration by parts.
  2. Choose uu and dvdv based on their differentiability and simplicity upon differentiation/integration.
  3. Apply the integration by parts formula.
  4. Simplify and integrate any remaining terms.
  5. Combine the results and add the constant of integration.

Reference

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