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

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(D) Let $I = \int \frac{dx}{\sin x + \cos x}$.Using the substitution $	an \frac{x}{2} = t$,we have $dx = \frac{2}{1+t^2} dt$,$\sin x = \frac{2t}{1+t^

Vedclass · Accepted Answer

$-\frac{1}{\sqrt{2}} \log \left(\frac{	an \frac{x}{2}-(\sqrt{2}+1)}{	an \frac{x}{2}+\sqrt{2}-1}ight)+c$,where $c$ is a constant of integration.

Vedclass · Answer

(D) Let $I = \int \frac{dx}{\sin x + \cos x}$.Using the substitution $	an \frac{x}{2} = t$,we have $dx = \frac{2}{1+t^2} dt$,$\sin x = \frac{2t}{1+t^2}$,and $\cos x = \frac{1-t^2}{1+t^2}$.Substituting these into the integral:$I = \int \frac{\frac{2}{1+t^2}}{\frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2}} dt = \int \frac{2}{2t + 1 - t^2} dt = -2 \int \frac{1}{t^2 - 2t - 1} dt$.Completing the square in the denominator:$t^2 - 2t - 1 = (t-1)^2 - 2 = (t-1)^2 - (\sqrt{2})^2$.Using the formula $\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} ight| + c$:$I = -2 	imes \frac{1}{2\sqrt{2}} \log \left| \frac{t-1-\sqrt{2}}{t-1+\sqrt{2}} ight| + c = -\frac{1}{\sqrt{2}} \log \left| \frac{	an \frac{x}{2} - 1 - \sqrt{2}}{	an \frac{x}{2} - 1 + \sqrt{2}} ight| + c$.Simplifying the expression:$I = -\frac{1}{\sqrt{2}} \log \left| \frac{	an \frac{x}{2} - (\sqrt{2} + 1)}{	an \frac{x}{2} + \sqrt{2} - 1} ight| + c$.

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

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