int frac{sin^2 x cos^2 x}{(sin^5 x + cos^3 x | vedclass

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

Question

(A) 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 + \

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) 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: $\sin^5 x + \cos^3 x \sin^2 x + \sin^3 x \cos^2 x + \cos^5 x = \sin^2 x(\sin^3 x + \cos^3 x) + \cos^2 x(\sin^3 x + \cos^3 x) = (\sin^2 x + \cos^2 x)(\sin^3 x + \cos^3 x) = (\sin^3 x + \cos^3 x)$. Thus,$I = \int \frac{\sin^2 x \cos^2 x}{(\sin^3 x + \cos^3 x)^2} dx$. Divide numerator and denominator by $\cos^6 x$: $I = \int \frac{	an^2 x \sec^2 x}{(	an^3 x + 1)^2} dx$. Let $t = 1 + 	an^3 x$,then $dt = 3 	an^2 x \sec^2 x dx$,so $	an^2 x \sec^2 x dx = \frac{dt}{3}$. $I = \frac{1}{3} \int \frac{dt}{t^2} = \frac{1}{3} (-t^{-1}) + C = \frac{-1}{3(1 + 	an^3 x)} + C$.

$\int \frac{\sin^2 x \cos^2 x}{(\sin^5 x + \cos^3 x \sin^2 x + \sin^3 x \cos^2 x + \cos^5 x)^2} dx$

Similar Questions

$\int(3-x) \sqrt{4-x} \, dx = $ (Where $C$ is a constant of integration.)

The value of $I=\int \frac{x^2}{(a+b x)^2} \,d x$ is

$\int {\frac{{10{x^9} + {{10}^x}{{\log }_e}10}}{{{{10}^x} + {x^{10}}}}} \;dx = $

$A$ primitive of $f(x) = \frac{x}{1 + x^2}$ is:

If $\int \frac{\sqrt{1-x^2}}{x^4} \,dx = A(x)\left(\sqrt{1-x^2}\right)^{m} + c$ for a suitably chosen integer $m$ and a function $A(x)$,where $c$ is a constant of integration,then $(A(x))^{m}$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module