Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Vanya Saxena Grade: Upto college level
        ∫√[sin(x-α)/sin(x+α)] dx
8 years ago

Jitender Singh
IIT Delhi
158 Points
										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 & RegardsJitender SinghIIT DelhiaskIITians Faculty

3 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Integral Calculus

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details