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Question Number 141847 by iloveisrael last updated on 24/May/21

$$\mathcal{I}=\int \frac{\sec x}{1+\csc x}dx$$

Answered by MJS_new last updated on 24/May/21

$$\int \frac{\sec x}{1+\csc x}=$$$$[t=\tan \frac{x}{2}\to dx={2\cos }^2\frac{x}{2}dt]$$$$=-4\int \frac{t}{(t-1){(t+1)}^3}dt=$$$$=-\frac{1}{2}\int \frac{dt}{t-1}-2\int \frac{dt}{{(t+1)}^3}+\int \frac{dt}{{(t+1)}^2}+\frac{1}{2}\int \frac{dt}{t+1}=$$$$=-\frac{1}{2}\ln (t-1)+\frac{1}{{(t+1)}^2}-\frac{1}{t+1}+\frac{1}{2}\ln (t+1)=$$$$=-\frac{t}{{(t+1)}^2}+\frac{1}{2}\ln \frac{t+1}{t-1}=...$$$$=-\frac{\sin x}{2(1+\sin x)}+\frac{1}{2}\ln \mid \frac{\cos x}{1-\sin x}\mid +C$$

Answered by iloveisrael last updated on 24/May/21

$$I=\int \frac{\frac{1}{\cos x}}{1+\frac{1}{\sin x}}dx=\int \frac{\sin x}{\cos x\sin x+\cos x}dx$$$$I=\int \frac{\sin x}{\cos x(\sin x+1)}dx$$$$I=\int \frac{\sin x(\sin x-1)}{-\cos {}^{3}x}dx$$$$I=-\int \frac{\sin {}^{2}x-\sin x}{\cos {}^{3}x}dx$$$$I=-\{\int \sec {}^{3}xdx-\int \sec xdx\}-$$$$\int \frac{d\left(\cos x\right)}{\cos {}^{3}x}$$$$I=\ln \mid \sec x+\tan x\mid +\frac{1}{2}\sec {}^{2}x-$$$$\int \sec {}^{3}xdx$$

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