int {frac{{dx}}{{{{cos }^3}xsqrt {2sin 2x} }} | vedclass

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

Question

(A) Let $I = \int {\frac{{dx}}{{{{\cos }^3}x\sqrt {2\sin 2x} }}} $.Using $\sin 2x = 2\sin x \cos x$,we get $I = \int {\frac{{dx}}{{{{\cos }^3}x\sqrt {

Vedclass · Accepted Answer

$\sqrt {	an x} + \frac{1}{5}{	an ^{5/2}}x + c$

Vedclass · Answer

(A) Let $I = \int {\frac{{dx}}{{{{\cos }^3}x\sqrt {2\sin 2x} }}} $.Using $\sin 2x = 2\sin x \cos x$,we get $I = \int {\frac{{dx}}{{{{\cos }^3}x\sqrt {4\sin x \cos x} }}} = \frac{1}{2}\int {\frac{{dx}}{{{{\cos }^{7/2}}x{{\sin }^{1/2}}x}}} $.This can be written as $I = \frac{1}{2}\int {\frac{{{{\sec }^4}x}}{{\sqrt {	an x} }}dx} = \frac{1}{2}\int {\frac{{(1 + {{	an }^2}x){{\sec }^2}x}}{{\sqrt {	an x} }}dx} $.Substitute $	an x = t$,so ${\sec ^2}x dx = dt$.$I = \frac{1}{2}\int {\frac{{1 + {t^2}}}{{\sqrt t }}dt} = \frac{1}{2}\int {({t^{ - 1/2}} + {t^{3/2}})dt} $.$I = \frac{1}{2} [2{t^{1/2}} + \frac{2}{5}{t^{5/2}}] + c = {t^{1/2}} + \frac{1}{5}{t^{5/2}} + c$.Substituting back $t = 	an x$,we get $I = \sqrt {	an x} + \frac{1}{5}{	an ^{5/2}}x + c$.

$\int {\frac{{dx}}{{{{\cos }^3}x\sqrt {2\sin 2x} }}} $ is equal to

Similar Questions

If $\int \frac{\sin \left(x-\frac{\pi}{4}\right)}{2+\sin 2 x} d x=-\frac{1}{\sqrt{2}} \tan ^{-1}(f(x))+C$,then $f(x)=$

If $\int \sqrt[3]{x}\left\{1+\sqrt[3]{x^4}\right\}^{1 / 7} d x=A\left(1+\sqrt[3]{x^4}\right)^B+c$,then the value of $A B$ is equal to

The integral $\int \frac{\sec^2 x}{(\sec x+\tan x)^{9/2}} dx$ equals (for some arbitrary constant $K$):

If $\int \frac{x^3 \, dx}{\sqrt{1+x^2}} = a(1+x^2) \sqrt{1+x^2} + b \sqrt{1+x^2} + c$ (where $c$ is a constant of integration),then the value of $3ab$ is

$\int x\sqrt{1 + x^2} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module