/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 32

Evaluate the following integrals. $$\int \tan ^{9} x \sec ^{4} x d x$$

Short Answer

Expert verified
The integral of this product is given by the expression: $$\int \tan ^{9} x \sec ^{4} x d x = \frac{1}{14\cos^{14} x} + C$$.

Step by step solution

01

Rewrite the integral using properties.

Recall that \(\tan x = \frac{\sin x}{\cos x}\) and \(\sec x = \frac{1}{\cos x}\). We can rewrite the given integral as follows: $$\int \tan ^{9} x \sec ^{4} x d x = \int \left(\frac{\sin x}{\cos x}\right)^9 \left(\frac{1}{\cos x}\right)^4 d x$$
02

Change of variable substituition.

Define \(u = \cos x\), then \(-du = \sin x\, dx\). Hence, $$\int \tan ^{9} x \sec ^{4} x d x = \int \left(\frac{-du}{u}\right) \left(\frac{1}{u^{14}}\right)$$ This gives us $$\int \tan ^{9} x \sec ^{4} x d x = -\int \frac{1}{u^{15}} du$$
03

Evaluate the integral using power rule.

Using the power rule for integrals, we have $$-\int \frac{1}{u^{15}} du = -\frac{u^{-14}}{(-14)} + C$$
04

Substitute back the original variable.

Replace \(u\) back with \(\cos x\). We get $$-\frac{u^{-14}}{(-14)} + C = \frac{1}{14\cos^{14} x} + C$$ So the solution is: $$\int \tan ^{9} x \sec ^{4} x d x = \frac{1}{14\cos^{14} x} + C$$.

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