Mathematics

# The integral $\displaystyle\int \dfrac{\sin^2x \cos^2x}{(\sin^5x+ \cos^3x \sin^2 x+ \sin^3x \cos^2x + \cos^5x)^2}dx$

$\dfrac{-1}{3(1+ \tan^3x)}+c$

##### SOLUTION
we have to evaluate,
$\Rightarrow I=\int\dfrac{sin^2x\ cos^2\ x}{(sin^5x+cos^3x\sin^2x+sin^3x\ cos^2x+cos^5x)^2}dx$

$\Rightarrow \int\dfrac{sin^2 x\ cos^2x}{\begin{Bmatrix}(sin^2x(sin^3x+ cos^3x)+cos^2x(sin^3x+ cos^3x)\end{Bmatrix}^2}dx$

$\Rightarrow \int\dfrac{sin^2 x\ cos^2x}{\begin{Bmatrix}(sin^2x+ cos^2x)(sin^3x+ cos^3x)\end{Bmatrix}^2}dx$

$\Rightarrow \int\dfrac{sin^2 x\ cos^2x}{\begin{Bmatrix}(sin^3x+ cos^3x)\end{Bmatrix}^2}dx$

$\Rightarrow \int\dfrac{sin^2 x\ cos^2x}{\begin{Bmatrix}sin^6x+2sin^3xcos^3x+ cos^6x\end{Bmatrix}}dx$

on dividing numerator and denominator by $cos^6x$we get

$\Rightarrow \int\dfrac{tan^2 x\ sec^2x}{\begin{Bmatrix}tan^6x+2tan^3x+1\end{Bmatrix}}dx$

$\Rightarrow \int\dfrac{tan^2 x\ sec^2x}{\begin{pmatrix}1+tan^3x\end{pmatrix}^2}dx$

Let $(1+ tan^3x )=t\Rightarrow 3tan^2x\ sec^2xdx =dt$

$\Rightarrow \dfrac13\int\dfrac{dt}{\begin{pmatrix}t\end{pmatrix}^2}= -\dfrac {1}{3}\ \dfrac1t + C$

$=\dfrac{-1}{3(1+ tan^3x)}+ C$

$\therefore \text{option A is correct}$

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Single Correct Medium Published on 17th 09, 2020
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