3.4 Q-2

Question Statement

Evaluate the following integrals:

i. $\int \tan ^{4} x , dx$

ii. $\int \sec^{4} x , dx$

iii. $\int e^{x} \sin 2x \cos 2x , dx$

iv. $\int \tan ^{3} x \sec x , dx$

v. $\int x^{3} e^{5x} , dx$

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Background and Explanation

In these types of integrals, you will use a variety of integration techniques such as:

• Power reduction: Reducing powers of trigonometric functions using trigonometric identities.
• Substitution: Using trigonometric and exponential substitutions to simplify the integrals.
• Integration by parts: A method used to integrate products of functions like $e^x$ and trigonometric functions.
• Standard Integrals: Utilizing known integral results for functions like $e^x$, $\sin x$, and $\cos x$.

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Solution

i. $\int \tan^{4} x , dx$

We begin by rewriting $\tan^4 x$ as $(\tan^2 x)^2$ to simplify the integral:

$$I = \int \tan^4 x , dx = \int \tan^2 x \cdot \tan^2 x , dx$$

Using the identity $\tan^2 x = \sec^2 x - 1$, we get:

$$I = \int (\sec^2 x - 1) \tan^2 x , dx$$

Breaking it into two integrals:

$$I = \int \sec^2 x \tan^2 x , dx - \int \tan^2 x , dx$$

The first integral is solved using substitution, and the second is a standard integral.

Final result:

$$I = \frac{\tan^3 x}{3} - \tan x + C$$

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ii. $\int \sec^{4} x , dx$

We start by using the identity $\sec^4 x = \sec^2 x \cdot \sec^2 x$. Then we expand and simplify:

$$I = \int \sec^2 x \cdot \sec^2 x , dx$$

Using the identity $\sec^2 x = 1 + \tan^2 x$, the integral becomes:

$$I = \int (1 + \tan^2 x) \sec^2 x , dx$$

Breaking this into two integrals:

$$I = \int \sec^2 x \tan^2 x , dx - \int \sec^2 x , dx$$

Solving each part, we get:

$$I = \frac{\tan^3 x}{3} + \tan x + C$$

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iii. $\int e^{x} \sin 2x \cos 2x , dx$

First, apply the product-to-sum identity for $\sin 2x \cos 2x$:

$$\sin 2x \cos 2x = \frac{1}{2} [\sin(3x) + \sin(x)]$$

The integral becomes:

$$I = \frac{1}{2} \int e^x [\sin 3x + \sin x] , dx$$

This is split into two integrals, $I_1$ and $I_2$:

$$I_1 = \int e^x \sin 3x , dx, \quad I_2 = \int e^x \sin x , dx$$

For both integrals, integration by parts is applied, leading to:

$$I = \frac{e^x}{4} \left[\frac{1}{5} \sin 3x - \frac{3}{5} \cos 3x + \sin x - \cos x \right] + C$$

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iv. $\int \tan^3 x \sec x , dx$

We rewrite $\tan^3 x \sec x$ as $\tan^2 x \cdot \tan x \sec x$ and use the identity $\tan^2 x = \sec^2 x - 1$:

$$I = \int (\sec^2 x - 1) \tan x \sec x , dx$$

This integral is split into two parts, and solving using substitution results in:

$$I = \frac{1}{3} [\sec x \tan^2 x - 2 \sec x] + C$$

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v. $\int x^3 e^{5x} , dx$

This is solved using integration by parts. Start by setting $u = x^3$ and $dv = e^{5x} dx$. Using integration by parts recursively:

$$I = \frac{e^{5x}}{5} x^3 - \frac{3}{5} \int e^{5x} x^2 , dx$$

Continuing the process until all powers of $x$ are reduced, we arrive at the final result:

$$I = \frac{e^{5x}}{5} x^3 - \frac{3}{25} e^{5x} x^2 + \dots + C$$

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Key Formulas or Methods Used

• Integration by Parts: $\int u , dv = uv - \int v , du$
• Power Reduction: $\tan^2 x = \sec^2 x - 1$
• Product-to-Sum Identity: $\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)]$
• Standard Integrals:
  • $\int e^{ax} , dx = \frac{e^{ax}}{a}$
  • $\int \sin x , dx = -\cos x$
  • $\int \cos x , dx = \sin x$

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Summary of Steps

1. For each integral, use trigonometric identities or substitution to simplify the integrals.
2. Apply appropriate methods like power reduction, substitution, or integration by parts.
3. For integrals involving exponentials and trigonometric functions, use standard integration techniques or recursion as necessary.
4. Simplify and combine the results to obtain the final answer.

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Reference

By Great Science Academy: