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∫√[sin(x-α)/sin(x+α)] dx

∫√[sin(x-α)/sin(x+α)] dx

Grade:Upto college level

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 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 )}}}dxI = \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
askIITians Faculty

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