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3.3 Q-19

Question Statement

Evaluate the following integral:

∫[cos⁑(xβˆ’x2)]Γ—[1xβˆ’1],dx\int \left[ \cos \left( \sqrt{x} - \frac{x}{2} \right) \right] \times \left[ \frac{1}{\sqrt{x}} - 1 \right] , dx

Background and Explanation

To solve this integral, we will use a substitution method to simplify the expression. The integrand consists of a cosine function multiplied by a rational expression. A useful substitution will make the cosine function easier to handle and transform the integral into a simpler form.


Solution

Let’s solve the integral step by step:

  1. Substitution: We begin by setting the expression inside the cosine function as a new variable:
xβˆ’x2=u \sqrt{x} - \frac{x}{2} = u

Now, differentiate both sides with respect to xx:

ddx(xβˆ’x2)=12xβˆ’12 \frac{d}{dx} \left( \sqrt{x} - \frac{x}{2} \right) = \frac{1}{2\sqrt{x}} - \frac{1}{2}

Therefore, the differential dxdx becomes:

dx=2,du dx = 2 , du
  1. Transform the Integral: Substituting u=xβˆ’x2u = \sqrt{x} - \frac{x}{2} into the original integral, we also replace the differential dxdx as 2,du2 , du. The term 1xβˆ’1\frac{1}{\sqrt{x}} - 1 becomes part of dudu, so we have:
∫[cos⁑(xβˆ’x2)]Γ—[1xβˆ’1],dx=∫cos⁑(u)Γ—2,du \int \left[ \cos \left( \sqrt{x} - \frac{x}{2} \right) \right] \times \left[ \frac{1}{\sqrt{x}} - 1 \right] , dx = \int \cos(u) \times 2 , du
  1. Simplify the Integral: The integral becomes:
2∫cos⁑(u),du 2 \int \cos(u) , du
  1. Integrate: The integral of cos⁑(u)\cos(u) is sin⁑(u)\sin(u), so we get:
2sin⁑(u)+C 2 \sin(u) + C
  1. Substitute Back: Finally, substitute back the expression for uu:
u=xβˆ’x2 u = \sqrt{x} - \frac{x}{2}

Therefore, the solution to the integral is:

2sin⁑(xβˆ’x2)+C 2 \sin \left( \sqrt{x} - \frac{x}{2} \right) + C

Key Formulas or Methods Used

  • Substitution Method: The substitution u=xβˆ’x2u = \sqrt{x} - \frac{x}{2} helped simplify the integral.
  • Basic Integration: The integral of cos⁑(u)\cos(u) is sin⁑(u)\sin(u).

Summary of Steps

  1. Substitute xβˆ’x2=u\sqrt{x} - \frac{x}{2} = u to simplify the expression.
  2. Differentiate to find dx=2,dudx = 2 , du.
  3. Substitute into the integral, resulting in 2∫cos⁑(u),du2 \int \cos(u) , du.
  4. Integrate to get 2sin⁑(u)+C2 \sin(u) + C.
  5. Substitute u=xβˆ’x2u = \sqrt{x} - \frac{x}{2} back into the solution:
2sin⁑(xβˆ’x2)+C 2 \sin \left( \sqrt{x} - \frac{x}{2} \right) + C