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

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(A) Let $I = \int \frac{d x}{\sin x + \cos x}$.Multiply and divide by $\sqrt{2}$:$I = \frac{1}{\sqrt{2}} \int \frac{d x}{\frac{1}{\sqrt{2}} \sin x + \

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}} \log \left|	an \left(\frac{x}{2}+\frac{\pi}{8}ight)ight|+C$

Vedclass · Answer

(A) Let $I = \int \frac{d x}{\sin x + \cos x}$.Multiply and divide by $\sqrt{2}$:$I = \frac{1}{\sqrt{2}} \int \frac{d x}{\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x}$.Using $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$, we get:$I = \frac{1}{\sqrt{2}} \int \frac{d x}{\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4}} = \frac{1}{\sqrt{2}} \int \frac{d x}{\sin \left(x + \frac{\pi}{4}ight)}$.Using $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$:$I = \frac{1}{\sqrt{2}} \int \frac{d x}{2 \sin \left(\frac{x}{2} + \frac{\pi}{8}ight) \cos \left(\frac{x}{2} + \frac{\pi}{8}ight)}$.Divide numerator and denominator by $\cos^2 \left(\frac{x}{2} + \frac{\pi}{8}ight)$:$I = \frac{1}{2 \sqrt{2}} \int \frac{\sec^2 \left(\frac{x}{2} + \frac{\pi}{8}ight) d x}{	an \left(\frac{x}{2} + \frac{\pi}{8}ight)}$.Let $t = 	an \left(\frac{x}{2} + \frac{\pi}{8}ight)$, then $dt = \frac{1}{2} \sec^2 \left(\frac{x}{2} + \frac{\pi}{8}ight) d x$, so $\sec^2 \left(\frac{x}{2} + \frac{\pi}{8}ight) d x = 2 dt$.$I = \frac{1}{2 \sqrt{2}} \int \frac{2 dt}{t} = \frac{1}{\sqrt{2}} \int \frac{dt}{t} = \frac{1}{\sqrt{2}} \log |t| + C$.Substituting $t$ back:$I = \frac{1}{\sqrt{2}} \log \left| 	an \left(\frac{x}{2} + \frac{\pi}{8}ight) ight| + C$.

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

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