The integral int frac{sin ^2 x cos ^2 x}{left | vedclass

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

Question

(B) Let $I = \int \frac{\sin ^2 x \cos ^2 x}{(\sin ^5 x + \cos ^3 x \sin ^2 x + \sin ^3 x \cos ^2 x + \cos ^5 x)^2} \,dx$. Factor the denominator: $\s

Vedclass · Accepted Answer

$\frac{-1}{3\left(1+	an ^3 xight)}+c$, where $c$ is a constant of integration.

Vedclass · Answer

(B) Let $I = \int \frac{\sin ^2 x \cos ^2 x}{(\sin ^5 x + \cos ^3 x \sin ^2 x + \sin ^3 x \cos ^2 x + \cos ^5 x)^2} \,dx$. Factor the denominator: $\sin ^5 x + \sin ^3 x \cos ^2 x + \cos ^3 x \sin ^2 x + \cos ^5 x = \sin ^3 x(\sin ^2 x + \cos ^2 x) + \cos ^3 x(\sin ^2 x + \cos ^2 x) = \sin ^3 x + \cos ^3 x$. Thus, $I = \int \frac{\sin ^2 x \cos ^2 x}{(\sin ^3 x + \cos ^3 x)^2} \,dx$. Divide the numerator and denominator by $\cos ^6 x$: $I = \int \frac{	an ^2 x \sec ^2 x}{(1 + 	an ^3 x)^2} \,dx$. Let $t = 1 + 	an ^3 x$. Then $dt = 3 	an ^2 x \sec ^2 x \,dx$, which implies $	an ^2 x \sec ^2 x \,dx = \frac{1}{3} dt$. Substituting these into the integral: $I = \frac{1}{3} \int \frac{1}{t^2} \,dt = \frac{1}{3} (-t^{-1}) + c = -\frac{1}{3t} + c$. Substituting $t$ back: $I = \frac{-1}{3(1 + 	an ^3 x)} + c$.

The integral $\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2} \,d x$ is equal to

Similar Questions

$\int \sqrt{\frac{\cos x - \cos^3 x}{1 - \cos^3 x}} \, dx$ is equal to

$\int \frac{e^x \, dx}{\sqrt{1 - e^{2x}}} = $

If $\int \frac{dx}{x (\sqrt{x^4 - 1})} = \frac{1}{k} \sec^{-1} (x^k)$,then the value of $k = $

Integrate the function: $\frac{\cos x}{\sqrt{4-\sin ^{2} x}}$

$\int \frac{(3 x-2) \tan \left(\sqrt{9 x^2-12 x+1}\right)}{\sqrt{9 x^2-12 x+1}} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module