Q. 3TF Trigonometric integrals: The int... [FREE SOLUTION]

Chapter 5: Q. 3TF (page 420)

Trigonometric integrals: The integrals that follow can be solved by using algebra to write the integrands in the form $f' (u (x)) u' (x)$ so that $u -$ substitution will apply.
Solve $鈭� \sec^{4} x \tan^{3} x d x$ by using the Pythagorean identity role="math" localid="1649174356968" $\tan^{2} x + 1 = \sec^{2} x$ to rewrite the integrand as $(\tan^{2} x + 1) \tan^{3} x \sec^{2} x$ and then applying substitution with $u = \tan x$ .

Short Answer

Expert verified

The value of integral is $鈭� \sec^{4} x \tan^{3} x d x = \frac{\sec^{6} x}{6} + \frac{\sec^{4} x}{4} + C$ .

Step by step solution

Step 1. Given Information

Solve $鈭� \sec^{4} x \tan^{3} x d x$ by using the Pythagorean identity $\tan^{2} x + 1 = \sec^{2} x$ to rewrite the integrand as $(\tan^{2} x + 1) \tan^{3} x \sec^{2} x$ and then applying substitution with $u = \tan x$ .

Step 2. The given integral is ∫sec4xtan3xdx

We can write as

$鈭� \sec^{4} x \tan^{3} x d x = 鈭� (\tan^{2} x + 1) \tan^{3} x \sec^{2} x d x$

Let

$u = \tan x \frac{d u}{d x} = \sec^{2} x d u = \sec^{2} x d x$

Step 3. Now the integral is

$鈭� \sec^{4} x \tan^{3} x d x = 鈭� (u^{2} + 1) u^{3} d u 鈭� \sec^{4} x \tan^{3} x d x = 鈭� (u^{2} 路 u^{3} + 1 路 u^{3}) d u 鈭� \sec^{4} x \tan^{3} x d x = 鈭� (u^{5} + u^{3}) d u 鈭� \sec^{4} x \tan^{3} x d x = 鈭� u^{5} d u + 鈭� u^{3} d u 鈭� \sec^{4} x \tan^{3} x d x = [\frac{u^{5 + 1}}{5 + 1}] + [\frac{u^{3 + 1}}{3 + 1}] + C 鈭� \sec^{4} x \tan^{3} x d x = \frac{u^{6}}{6} + \frac{u^{4}}{4} + C 鈭� \sec^{4} x \tan^{3} x d x = \frac{\sec^{6} x}{6} + \frac{\sec^{4} x}{4} + C$

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Short Answer

Step by step solution

Step 1. Given Information

Step 2. The given integral is ∫sec4xtan3xdx

Step 3. Now the integral is

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