# Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 403: 32

$$\frac{1}{3}\cos^3x-3\cos x-3\sec x+\frac{1}{3}\sec^3x +c$$

#### Work Step by Step

\begin{align*} \int \frac{\sin ^{7} x} {\cos ^{4} x}d x&= \int \sin^{6}x\cos^{-4}x\sin xdx\\ &= \int \cos^{-4}x(1-\cos^2x)^3\sin xdx\\ &=\int \cos^{-4}x(-\cos^6x+3\cos^4x-3\cos^2x+1 )\sin xdx\\ &= \int \left( -\cos^2x+3-3\cos^{-2}x+\cos^{-4}x \right)\sin xdx\\ &= \frac{1}{3}\cos^3x-3\cos x-3\sec x+\frac{1}{3}\sec^3x +c \end{align*}

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