The integral int frac{sin^2 x cos^2 x}{(sin^3 | vedclass

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

Question

(B) Let $I = \int \frac{\sin^2 x \cos^2 x}{(\sin^3 x + \cos^3 x)^2} dx$Divide numerator and denominator by $\cos^6 x$:$I = \int \frac{\frac{\sin^2 x \

Vedclass · Accepted Answer

$-\frac{1}{3(1 + 	an^3 x)} + c$

Vedclass · Answer

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

The integral $\int \frac{\sin^2 x \cos^2 x}{(\sin^3 x + \cos^3 x)^2} dx$ is equal to

Similar Questions

Let $I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x$. If $I(0)=3$,then $I\left(\frac{\pi}{12}\right)$ is equal to :

Integrate the function: $\frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right]}{x^{4}}$

If $\int {\frac{{\left( {2x + 3} \right)dx}}{{x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) + 1}}} = C - \frac{1}{{f(x)}}$ where $f(x)$ is of the form of $ax^2 + bx + c$ then $(a + b + c)$ equals

$\int \frac{\cos x + x \sin x}{x(x - \cos x)} dx = $

$\int \frac{dx}{\sqrt{x-x^2}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module