int frac{x+cos x}{1-sin x} d x= | vedclass

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Question

(A) We have $I = \int \frac{x+\cos x}{1-\sin x} dx$. Using the identities $\cos x = \sin(\frac{\pi}{2}-x)$ and $1-\sin x = 1-\cos(\frac{\pi}{2}-x) = 2

Vedclass · Accepted Answer

$x 	an \left(\frac{\pi}{4}+\frac{x}{2}ight)+c$

Vedclass · Answer

(A) We have $I = \int \frac{x+\cos x}{1-\sin x} dx$. Using the identities $\cos x = \sin(\frac{\pi}{2}-x)$ and $1-\sin x = 1-\cos(\frac{\pi}{2}-x) = 2\sin^2(\frac{\pi}{4}-\frac{x}{2})$, we rewrite the integral. Alternatively, note that $\frac{1}{1-\sin x} = \frac{1+\sin x}{\cos^2 x} = \sec^2 x + \sec x 	an x$. Then $I = \int x(\sec^2 x + \sec x 	an x) dx + \int \frac{\cos x}{1-\sin x} dx$. Using integration by parts on the first term: $u=x, dv=(\sec^2 x + \sec x 	an x)dx \implies v=	an x + \sec x$. $I = x(	an x + \sec x) - \int (	an x + \sec x) dx + \int \frac{\cos x}{1-\sin x} dx$. Since $\int \frac{\cos x}{1-\sin x} dx = -\ln|1-\sin x|$, this approach is complex. Let's use the identity $\frac{1}{1-\sin x} = \sec^2(\frac{\pi}{4}+\frac{x}{2})$. Then $I = \int x \sec^2(\frac{\pi}{4}+\frac{x}{2}) dx + \int \frac{\cos x}{1-\sin x} dx$. Integrating by parts: $I = x \cdot 2 	an(\frac{\pi}{4}+\frac{x}{2}) - \int 2 	an(\frac{\pi}{4}+\frac{x}{2}) dx + \int \frac{\cos x}{1-\sin x} dx$. Simplifying, we get $I = x 	an(\frac{\pi}{4}+\frac{x}{2}) + c$.

$\int \frac{x+\cos x}{1-\sin x} d x=$

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