Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises - Page 484: 15

Answer

$$\int\cot x\cos^2 xdx=\ln|\sin x|-\frac{\sin^2 x}{2}+C$$

Work Step by Step

$$A=\int\cot x\cos^2 xdx$$ Here we would try to change the whole integral in terms of $\sin x$ and $\cos x$ so that we can use the Substitution Rule, according to the following formulas $$\cot x=\frac{\cos x}{\sin x}$$ $$\cos^2 x=1-\sin^2 x$$ Therefore we have $$A=\int\frac{\cos x}{\sin x}(1-\sin^2 x)dx$$ $$A=\int\frac{\cos x}{\sin x}dx-\int\cos x\sin xdx$$ $$A=X-Y$$ *Consider X: $$X=\int\frac{\cos x}{\sin x}dx$$ Let $u=\sin x$. Then $du=\cos xdx$ $$X=\int \frac{1}{u}du$$ $$X=\ln|u|+C$$ $$X=\ln|\sin x|+C$$ Consider Y: $$C=\int\cos x\sin xdx$$ Again, let $v=\sin x$. Then $dv=\cos xdx$ $$Y=\int vdv$$ $$Y=\frac{v^2}{2}+C$$ $$Y=\frac{\sin^2 x}{2}+C$$ Combine X and Y, we have $$A=\ln|\sin x|-\frac{\sin^2 x}{2}+C$$

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