int frac{sqrt{cos 2 x}}{sin x} d x= | vedclass

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(B) Let $I = \int \frac{\sqrt{\cos 2x}}{\sin x} dx$.We know that $\cos 2x = \frac{1-	an^2 x}{1+	an^2 x}$.So, $I = \int \frac{\sqrt{1-	an^2 x}}{\sec

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}} \log \left|\frac{\sqrt{2}+\sqrt{1-	an ^2 x}}{\sqrt{2}-\sqrt{1-	an ^2 x}}ight|-\frac{1}{2} \log \left|\frac{1+\sqrt{1-	an ^2 x}}{1-\sqrt{1-	an ^2 x}}ight|+c$

Vedclass · Answer

(B) Let $I = \int \frac{\sqrt{\cos 2x}}{\sin x} dx$.We know that $\cos 2x = \frac{1-	an^2 x}{1+	an^2 x}$.So, $I = \int \frac{\sqrt{1-	an^2 x}}{\sec x \sin x} dx = \int \frac{\sqrt{1-	an^2 x}}{	an x} dx$.Let $1-	an^2 x = t^2$, then $-2	an x \sec^2 x dx = 2t dt$, which implies $dx = \frac{-t dt}{	an x (1+	an^2 x)} = \frac{-t dt}{\sqrt{1-t^2}(2-t^2)}$.Substituting these, $I = \int \frac{t}{\sqrt{1-t^2}} \cdot \frac{-t dt}{\sqrt{1-t^2}(2-t^2)} = -\int \frac{t^2}{(1-t^2)(2-t^2)} dt$.Using partial fractions, $\frac{t^2}{(1-t^2)(2-t^2)} = \frac{2}{t^2-2} - \frac{1}{t^2-1}$.Thus, $I = -\int \left( \frac{2}{t^2-2} - \frac{1}{t^2-1} ight) dt = 2 \int \frac{1}{2-t^2} dt + \int \frac{1}{t^2-1} dt$.Using standard integrals, $I = 2 \cdot \frac{1}{2\sqrt{2}} \log \left| \frac{\sqrt{2}+t}{\sqrt{2}-t} ight| + \frac{1}{2} \log \left| \frac{t-1}{t+1} ight| + c$.Substituting $t = \sqrt{1-	an^2 x}$, we get $I = \frac{1}{\sqrt{2}} \log \left| \frac{\sqrt{2}+\sqrt{1-	an^2 x}}{\sqrt{2}-\sqrt{1-	an^2 x}} ight| - \frac{1}{2} \log \left| \frac{1+\sqrt{1-	an^2 x}}{1-\sqrt{1-	an^2 x}} ight| + c$.

$\int \frac{\sqrt{\cos 2 x}}{\sin x} d x=$

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