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Integral Calculus> indefinite integration...

question mark

∫√[sin(x-α)/sin(x+α)] dx

Vanya Saxena , 15 Years ago

Grade Upto college level

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:

$I = \int \sqrt{\frac{sin(x-\alpha )}{sin(x+\alpha )}}dx$

$I = \int {\frac{sin(x-\alpha )}{\sqrt{sin(x+\alpha).sin(x-\alpha )}}}dx$ $I = \int {\frac{sin(x).cos\alpha -cos(x)sin\alpha }{\sqrt{sin^{2}(x)-sin^{2}(\alpha )}}}dx$

$I = \int \frac{sin(x).cos\alpha }{\sqrt{cos^{2}\alpha -cos^{2}(x)}}dx + \int \frac{-cos(x).sin\alpha }{\sqrt{sin^{2}(x)-sin^{2}\alpha }}dx$

$I_{1} = \int \frac{sin(x).cos\alpha }{\sqrt{cos^{2}\alpha -cos^{2}(x)}}dx$

$cosx = t$

$-sinx = dt$

$I_{1} = \int \frac{-cos\alpha }{\sqrt{cos^{2}\alpha -t^{2}}}dt$

$I_{1} =-cos\alpha.sin^{-1}(\frac{t}{cos\alpha }) + constant$

$I_{1} =-cos\alpha.sin^{-1}(\frac{cosx}{cos\alpha }) + constant$

$I_{2} = \int \frac{-cos(x).sin\alpha }{\sqrt{sin^{2}(x)-sin^{2}\alpha }}dx$

$sinx = t$

$cosx = dt$

$I_{2} = \int \frac{-sin\alpha }{\sqrt{t^{2}-sin^{2}\alpha }}dt$

$I_{2} = -sin\alpha.ln|t + \sqrt{t^{2}-sin^{2}\alpha } + constant$

$I_{2} = -sin\alpha.ln|sin(x) + \sqrt{sin^{2}(x)-sin^{2}\alpha }| + constant$

$I = I_{1}+I_{2}$

$I =-cos\alpha.sin^{-1}(\frac{cosx}{cos\alpha }) -sin\alpha.ln|sin(x) + \sqrt{sin^{2}(x)-sin^{2}\alpha }| + constant$

Thanks & Regards

Jitender Singh

IIT Delhi

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