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S Suraj Bhandari
Chain Rule is one of the most important concepts in Differentiations That's why we will discuss it in upcoming concepts till then you can take some basic idea of this concept Thanking You !!

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H Himanshu Meshram

@Vinod

\\\int \sqrt{\frac{\sin \left(x-a\right)}{\sin \left(x+a\right)}}dx\\rationalize\;it\\\int \sqrt{\frac{\sin \left(x-a\right)\sin \left(x-a\right)}{\sin \left(x+a\right)\sin \left(x-a\right)}}dx\\\int \frac{\sin \left(x-a\right)}{\sqrt{\sin \:\left(x+a\right)\sin \:\left(x-a\right)}}dx\\\because \sin \left(A+B\right)\sin \left(A-B\right)=\sin ^2A-\sin ^2B\\\int \frac{\sin x\:\cos a-\sin a\:\cos x}{\sqrt{\sin ^2x-\sin ^2a}}dx\\\cos a\int \frac{\sin x\:}{\sqrt{\sin ^2x-\sin ^2a}}dx\:-\sin a\int \:\frac{\:\cos \:x}{\sqrt{\sin \:^2x-\sin \:^2a}}dx\:\:\:

\\\cos a\int \frac{\sin x\:}{\sqrt{\sin ^2x-\sin ^2a}}dx\:\\\cos \:a\int \frac{\sin \:x\:}{\sqrt{1-\cos ^2x-1+\cos ^2a}}dx\:\\\Rightarrow \cos \:a\int \frac{\sin \:x\:}{\sqrt{\cos \:^2a-\cos ^2x}}dx\:\\put\;\cos x=t,and\;solve\\same\;method\;apply\;for\;\sin a\int \:\frac{\:\cos \:x}{\sqrt{\sin \:^2x-\sin \:^2a}}dx\:\:\:

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