## Calculus 10th Edition

Published by Brooks Cole

# Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.3 Exercises - Page 530: 24

#### Answer

$\frac{1}{2\pi} \tan^4\frac{\pi x}{2} +C$

#### Work Step by Step

Find the indefinite integral $\int tan^3\frac{\pi x}{2}sec^2\frac{\pi x}{2}dx$ Let $u=tan\frac{\pi x}{2}$, $du=\frac{\pi}{2}sec^2\frac{\pi x}{2}dx$ $\frac{2}{\pi}\int u^3du$, Integrate $\frac{1}{2 \pi}u^4 +C$, Resubstitute $\frac{1}{2\pi} tan^4\frac{\pi x}{2} +C$

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