Flag Integral Calculus> indefinite integration...
question mark

∫√[sin(x-α)/sin(x+α)] dx

Vanya Saxena , 15 Years ago
Grade Upto college level
anser 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 )}}}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

star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments