3.4 Q-5

Question Statement

Evaluate the following integrals:

1. $\int e^x \left( \frac{1}{x} + \ln x \right) dx$

2. $\int e^x (\cos x + \sin x) dx$

3. $\int e^{ax} \left[ \sec^{-1} x + \frac{1}{x \sqrt{x^2 - 1}} \right] dx$

4. $\int e^{3x}\left(\frac{3 \sin x - \cos x}{\sin^2 x}\right) dx$

5. $\int e^{2x}(-\sin x + 2 \cos x) dx$

6. $\int \frac{x e^x}{(1+x)^2} dx$

7. $\int e^{-x}[\cos x - \sin x] dx$

8. $\int \frac{e^m \tan^{-1}(x)}{1 + x^2} , dx$

9. $\int \frac{2x}{1 - \sin(x)} , dx$

10. $\int \frac{e^x(1 + x)}{(2 + x)} , dx$

11. $\int \frac{1 - \sin(x)}{1 - \cos(x)} e^x , dx$

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Background and Explanation

To solve these integrals, we will apply techniques like integration by parts and recognize patterns that align with derivative forms of common functions. The strategy often involves identifying functions whose derivatives match parts of the integrand. We will use the fact that:

• The integral of the derivative of a product can simplify the problem (integration by parts).
• Derivatives of logarithmic and inverse trigonometric functions are also essential tools in solving such integrals.

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Solution

i. $\int e^x \left( \frac{1}{x} + \ln x \right) dx$

• Recognize that we can split the integral: $\int e^x \left( \frac{1}{x} + \ln x \right) dx = \int e^x \cdot \frac{1}{x} dx + \int e^x \ln x dx$

• For the first part, apply integration by parts where: $u = \frac{1}{x}, , dv = e^x dx$

  This gives: $\int e^x \frac{1}{x} dx = e^x \ln x - \int e^x \ln x dx$

• The second part simplifies because of the known formula for integration by parts: $\int e^x \ln x dx = e^x \ln x - \int e^x \frac{1}{x} dx$

• Combining these, we find: $\int e^x \left( \frac{1}{x} + \ln x \right) dx = e^x \ln x + C$

ii. $\int e^x (\cos x + \sin x) dx$

• We observe that the integrand is the sum of $e^x$ multiplied by sine and cosine functions. Use integration by parts, recognizing that: $f(x) = \sin x, \quad f'(x) = \cos x$

• Apply the integration by parts formula: $\int e^x (\cos x + \sin x) dx = \int e^x \left[ f(x) + f'(x) \right] dx$

• This simplifies to: $\int e^x \left[ f(x) + f'(x) \right] dx = e^x \sin x + C$

iii. $\int e^{ax} \left[ \sec^{-1} x + \frac{1}{x \sqrt{x^2 - 1}} \right] dx$

• For the inverse secant function, recall that: $\frac{d}{dx} \sec^{-1} x = \frac{1}{x \sqrt{x^2 - 1}}$

• This allows us to simplify the integral as: $\int e^{ax} \left[ \sec^{-1} x + \frac{1}{x \sqrt{x^2 - 1}} \right] dx = \int e^{ax} \frac{d}{dx} \left( e^{ax} \sec^{-1} x \right) dx$

• Applying the fundamental theorem of calculus, we find: $e^{ax} \sec^{-1} x + C$

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iv. $\quad \int \mathrm{e}^{3 \mathrm{x}}\left(\frac{3 \operatorname{Sin} \mathrm{x}-\operatorname{Cos} \mathrm{x}}{\operatorname{Sin}^{2} \mathrm{x}}\right) dx$

We begin by breaking down the integral into two parts:

$$\int \mathrm{e}^{3 \mathrm{x}} \left( 3 \csc x - \cot x \csc x \right) dx$$

Apply integration by parts on both integrals.

1. First part:
   The first term can be integrated by taking $\csc x$ as the first function.
   $\int \mathrm{e}^{3 \mathrm{x}} \csc x , dx = \mathrm{e}^{3 \mathrm{x}} \csc x$

2. Second part:
   Similarly, integrate the second term using $\cot x \csc x$ as the first function.
   $\int \mathrm{e}^{3 \mathrm{x}} \cot x \csc x , dx$
   This yields:
   $\mathrm{e}^{3 \mathrm{x}} \cot x \csc x$

Thus, the final result is:

$$\boxed{\mathrm{e}^{3 \mathrm{x}} \cot x \csc x + C}$$

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v. $\int \mathrm{e}^{2 \mathrm{x}}(-\operatorname{Sin} \mathrm{x}+2 \operatorname{Cos} \mathrm{x}) dx$

We can split the integral into two parts:

$$2 \int \mathrm{e}^{2 \mathrm{x}} \cos x , dx - \int \mathrm{e}^{2 \mathrm{x}} \sin x , dx$$

1. First part:
   Use integration by parts on the first integral. Let $\cos x$ be the first function: $\int \mathrm{e}^{2 \mathrm{x}} \cos x , dx$
   This yields:
   $\mathrm{e}^{2 \mathrm{x}} \cos x$

2. Second part:
   The second integral, with $\sin x$ as the first function, results in: $\int \mathrm{e}^{2 \mathrm{x}} \sin x , dx$
   This yields:
   $\mathrm{e}^{2 \mathrm{x}} \sin x$

Thus, the final result is:

$$\boxed{\mathrm{e}^{2 \mathrm{x}} \cos x + C}$$

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vi. $\int \frac{x \mathrm{e}^{\mathrm{x}}}{(1+x)^{2}} dx$

We rewrite the integrand using the identity:

$$\int \mathrm{e}^{\mathrm{x}} \left( \frac{1}{1+x} - \frac{1}{(1+x)^2} \right) dx$$

Let $f(x) = \frac{1}{1+x}$. We know that:

$$f'(x) = -\frac{1}{(1+x)^2}$$

Thus, the integral becomes:

$$\int \mathrm{e}^{\mathrm{x}} \left[ f(x) + f'(x) \right] dx$$

This simplifies to:

$$\mathrm{e}^{\mathrm{x}} f(x) + C = \mathrm{e}^{\mathrm{x}} \frac{1}{1+x} + C$$

Thus, the final result is:

$$\boxed{\mathrm{e}^{\mathrm{x}} \frac{1}{1+x} + C}$$

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vii. $\int e^{-x}[\operatorname{Cos} x-\operatorname{Sin} x] dx$

We split the integral into two parts:

$$\int \mathrm{e}^{-x} \cos x , dx - \int \mathrm{e}^{-x} \sin x , dx$$

1. First part:
   Apply integration by parts to the first integral. Let $\cos x$ be the first function: $\int \mathrm{e}^{-x} \cos x , dx$
   This yields: $\mathrm{e}^{-x} \cos x$

2. Second part:
   Similarly, for the second integral, with $\sin x$ as the first function: $\int \mathrm{e}^{-x} \sin x , dx$
   This yields: $\mathrm{e}^{-x} \sin x$

Thus, the final result is:

$$\boxed{\mathrm{e}^{-x} (\cos x - \sin x) + C}$$

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viii. $\int \frac{e^m \tan^{-1}(x)}{1 + x^2} , dx$

Solution:

We start by using the substitution $y = \tan^{-1}(x)$. The derivative of this is:

$$\frac{dy}{dx} = \frac{1}{1 + x^2}$$

This implies:

$$dy = \frac{1}{1 + x^2} , dx$$

Thus, the integral becomes:

$$\int \frac{e^m \tan^{-1}(x)}{1 + x^2} , dx = \int e^m \cdot \tan^{-1}(x) \left(\frac{1}{1 + x^2}\right) dx$$

Since $\frac{1}{1 + x^2} , dx = dy$, the integral simplifies to:

$$\int e^m , dy$$

Now, integrating with respect to $y$:

$$= e^m \cdot y + C$$

Finally, substituting back $y = \tan^{-1}(x)$:

$$= \frac{e^m \tan^{-1}(x)}{m} + C$$

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ix. $\int \frac{2x}{1 - \sin(x)} , dx$

Solution:

Start by rewriting the expression:

$$\int \frac{2x}{1 - \cos\left(\frac{\pi}{2} - x\right)} , dx$$

We use a trigonometric identity to rewrite $1 - \sin^2\left(\frac{\pi}{2} - \frac{x}{2}\right)$, and simplify the expression:

$$= \int x \csc^2\left(\frac{\pi}{2} - \frac{x}{2}\right) , dx$$

Now, let’s use a substitution:

$$u = \frac{\pi}{2} - \frac{x}{2}$$

So:

$$- \frac{1}{2} dx = du \quad \text{or} \quad dx = -2 du$$

This changes the integral to:

$$\int \left(\frac{\pi}{2} - 2u\right) \csc^2(u) , (-2 du)$$

Expanding:

$$= -2 \int \left(\frac{\pi}{2} - 2u\right) \csc^2(u) , du$$

Now, we integrate:

$$= 2 \left[ -\cot(u)\left(\frac{\pi}{2} - 2u\right) + 4 \ln|\sin(u)| \right] + C$$

Finally, substituting back $u = \frac{\pi}{2} - \frac{x}{2}$:

$$= 2 \cot\left(\frac{\pi}{4} - \frac{x}{2}\right) + 4 \ln \left|\sin\left(\frac{\pi}{4} - \frac{x}{2}\right)\right| + C$$

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x. $\int \frac{e^x(1 + x)}{(2 + x)} , dx$

Solution:

First, simplify the integrand:

$$= \int \frac{e^x (1 + x - 1)}{(2 + x)^2} , dx$$

Now, rewrite it as:

$$= \int e^x \left[\frac{2 + x}{(2 + x)^2} - \frac{1}{(2 + x)^2}\right] , dx$$

Breaking it down:

$$= \int e^x \left[\frac{1}{2 + x} - \frac{1}{(2 + x)^2}\right] , dx$$

Now, focus on the first part $\frac{1}{2 + x}$, and recall that:

$$f(x) = \frac{1}{2 + x}, \quad f'(x) = -\frac{1}{(2 + x)^2}$$

Thus, we can write the integral as:

$$= \int e^x \left[ f(x) + f'(x) \right] , dx$$

Integrating this:

$$= e^x \cdot f(x) + C = e^x \cdot \frac{1}{2 + x} + C$$

Thus, the final result is:

$$= \frac{e^x}{2 + x} + C$$

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xi. $\int \frac{1 - \sin(x)}{1 - \cos(x)} e^x , dx$

Solution:

Start by simplifying the integrand:

$$= \int \left(\frac{1 - \sin(x)}{2 \sin\left(\frac{x}{2}\right)}\right) e^x , dx$$

Using a trigonometric identity, we express the terms involving sine and cosine in a simpler form:

$$= \int e^x \left[\frac{1}{2 \sin^2\left(\frac{x}{2}\right)} - \frac{\sin(x)}{2 \sin^2\left(\frac{x}{2}\right)}\right] , dx$$

This simplifies to:

$$= \int e^x \left[- \cot\left(\frac{x}{2}\right) - \csc^2\left(\frac{x}{2}\right)\right] , dx$$

Let $f(x) = -\cot\left(\frac{x}{2}\right)$. Thus, $f'(x) = \frac{1}{2} \csc^2\left(\frac{x}{2}\right)$, and we can write:

$$= -\int e^x \left[ f(x) + f'(x) \right] , dx$$

Integrating:

$$= -e^x \cdot f(x) + C = -e^x \cot\left(\frac{x}{2}\right) + C$$

Thus, the final result is:

$$= -e^x \cot\left(\frac{x}{2}\right) + C$$

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Summary of Steps

1. Break down the integral into simpler parts.
2. Apply integration by parts where applicable.
3. For complex expressions, identify known identities or use algebraic manipulation.
4. Apply the result to each part of the integral.
5. Simplify the final expression to get the solution.

Key Formulas or Methods Used

• Integration by Parts: $\int u dv = uv - \int v du$

• Derivative of the Inverse Secant Function: $\frac{d}{dx} \sec^{-1} x = \frac{1}{x \sqrt{x^2 - 1}}$

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Reference

By Great Science Academy: