int frac{sin x}{sin left(x-frac{pi}{4}right)} | vedclass

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Question

(A) Let $I = \int \frac{\sin x}{\sin \left(x-\frac{\pi}{4}\right)} d x$.Substitute $x = (x - \frac{\pi}{4}) + \frac{\pi}{4}$,so $I = \int \frac{\sin \

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}\left[x+\log \left|\sin \left(x-\frac{\pi}{4}\right)\right|\right]+c$

Vedclass · Answer

(A) Let $I = \int \frac{\sin x}{\sin \left(x-\frac{\pi}{4}\right)} d x$.Substitute $x = (x - \frac{\pi}{4}) + \frac{\pi}{4}$,so $I = \int \frac{\sin \left((x-\frac{\pi}{4}) + \frac{\pi}{4}\right)}{\sin \left(x-\frac{\pi}{4}\right)} d x$.Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get $I = \int \frac{\sin (x-\frac{\pi}{4}) \cos \frac{\pi}{4} + \cos (x-\frac{\pi}{4}) \sin \frac{\pi}{4}}{\sin (x-\frac{\pi}{4})} d x$.Since $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,we have $I = \int \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cot (x-\frac{\pi}{4}) \right) d x$.Integrating term by term,$I = \frac{1}{\sqrt{2}} \int 1 d x + \frac{1}{\sqrt{2}} \int \cot (x-\frac{\pi}{4}) d x$.The integral of $\cot u$ is $\log |\sin u|$,so $I = \frac{1}{\sqrt{2}} [x + \log |\sin (x-\frac{\pi}{4})|] + c$.

$\int \frac{\sin x}{\sin \left(x-\frac{\pi}{4}\right)} d x=$

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