∫√[sin(x-α)/sin(x+α)] dx

Jitender Singh IIT Delhi
9 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