int frac{dx}{cos x - sin x} is equal to | vedclass

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

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$\frac{1}{\sqrt{2}} \log \left| 	an \left( \frac{x}{2} + \frac{3\pi}{8} ight) ight| + c$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{\cos x - \sin x}$.Multiply and divide by $\sqrt{2}$:$I = \frac{1}{\sqrt{2}} \int \frac{dx}{\frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x}$Using $\cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,we get:$I = \frac{1}{\sqrt{2}} \int \frac{dx}{\cos x \cos \frac{\pi}{4} - \sin x \sin \frac{\pi}{4}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\cos(x + \frac{\pi}{4})}$$I = \frac{1}{\sqrt{2}} \int \sec(x + \frac{\pi}{4}) dx$Using the formula $\int \sec 	heta d	heta = \log |	an(\frac{	heta}{2} + \frac{\pi}{4})| + c$:$I = \frac{1}{\sqrt{2}} \log \left| 	an \left( \frac{x + \frac{\pi}{4}}{2} + \frac{\pi}{4} ight) ight| + c$$I = \frac{1}{\sqrt{2}} \log \left| 	an \left( \frac{x}{2} + \frac{\pi}{8} + \frac{2\pi}{8} ight) ight| + c$$I = \frac{1}{\sqrt{2}} \log \left| 	an \left( \frac{x}{2} + \frac{3\pi}{8} ight) ight| + c$.

$\int \frac{dx}{\cos x - \sin x}$ is equal to

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