int frac{1}{cos ^3 x sqrt{sin 2 x}} ,d x= | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int \frac{1}{\cos ^3 x \sqrt{\sin 2 x}} \,d x$.Since $\sin 2x = 2 \sin x \cos x$,we have $I = \int \frac{1}{\cos ^3 x \sqrt{2 \sin x \co

Vedclass · Accepted Answer

$\sqrt{2}\left(\sqrt{	an x}+\frac{1}{5}(	an x)^{\frac{5}{2}}ight)+c$,where $c$ is a constant of integration.

Vedclass · Answer

(A) Let $I = \int \frac{1}{\cos ^3 x \sqrt{\sin 2 x}} \,d x$.Since $\sin 2x = 2 \sin x \cos x$,we have $I = \int \frac{1}{\cos ^3 x \sqrt{2 \sin x \cos x}} \,d x = \frac{1}{\sqrt{2}} \int \frac{1}{\cos ^3 x \sqrt{\sin x} \sqrt{\cos x}} \,d x$.This simplifies to $I = \frac{1}{\sqrt{2}} \int \frac{1}{\cos^{3.5} x \sqrt{\sin x}} \,d x = \frac{1}{\sqrt{2}} \int \frac{\sec^{3.5} x}{\sqrt{	an x}} \,d x$.Wait,let us rewrite the integrand as $\frac{\sec^2 x \cdot \sec^2 x}{\sqrt{2 \sin x \cos x}} = \frac{\sec^2 x \cdot \sec^2 x}{\sqrt{2 	an x \cos^2 x}} = \frac{\sec^4 x}{\sqrt{2 	an x}}$.So,$I = \frac{1}{\sqrt{2}} \int \frac{\sec^4 x}{\sqrt{	an x}} \,d x = \frac{1}{\sqrt{2}} \int \frac{(1+	an^2 x) \sec^2 x}{\sqrt{	an x}} \,d x$.Let $	an x = t$,then $\sec^2 x \,d x = dt$.$I = \frac{1}{\sqrt{2}} \int \frac{1+t^2}{\sqrt{t}} \,dt = \frac{1}{\sqrt{2}} \int (t^{-1/2} + t^{3/2}) \,dt$.$I = \frac{1}{\sqrt{2}} \left( \frac{t^{1/2}}{1/2} + \frac{t^{5/2}}{5/2} ight) + c = \frac{1}{\sqrt{2}} \left( 2\sqrt{t} + \frac{2}{5} t^{5/2} ight) + c$.$I = \sqrt{2} \sqrt{t} + \frac{\sqrt{2}}{5} t^{5/2} + c = \sqrt{2} \left( \sqrt{	an x} + \frac{1}{5} (	an x)^{5/2} ight) + c$.

$\int \frac{1}{\cos ^3 x \sqrt{\sin 2 x}} \,d x=$

Similar Questions

If $\int \frac{x^{49} \tan ^{-1}(x^{50})}{1+x^{100}} d x=k(\tan ^{-1}(x^{50}))^2+c$, then $k=$

$\int \frac{\log (x + \sqrt {1 + x^2})}{\sqrt {1 + x^2}} \, dx = $

$\int \frac{f(x) g^{\prime}(x)-f^{\prime}(x) g(x)}{f(x) g(x)} \times [\log g(x)-\log f(x)] \, dx$ is equal to

$\int \sin \sqrt{x} \, dx = $

$\int \frac{1}{x^2 \sqrt{1-x^2}} \cdot d x = \dots + C$. Where,$(0 < |x| < 1)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module