Question

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

• A. $\frac{-1}{3(1+{\tan }^{3}x)}+c$
• B. $\frac{1}{1+{\cot }^{3}x}+c$
• C. $\frac{-1}{1+{\cot }^{3}x}+c$
• D. $\frac{1}{3(1+{\tan }^{3}x)}+c$

Answer 1

The correct option is A $\frac{-1}{3(1+{\tan }^{3}x)}+c$

we have to evaluate,

$\Rightarrow I=\int \frac{si{n}^{2}xco{s}^{2}x}{(si{n}^{5}x+co{s}^{3}x{\sin }^{2}x+si{n}^{3}xco{s}^{2}x+co{s}^{5}x{)}^2}dx$

$\Rightarrow \int \frac{si{n}^{2}xco{s}^{2}x}{{\left\{\begin{array}{c}\left(si{n}^{2}x\right(si{n}^{3}x+co{s}^{3}x)+co{s}^{2}x(si{n}^{3}x+co{s}^{3}x)\end{array}\right\}}^2}dx$

$\Rightarrow \int \frac{si{n}^{2}xco{s}^{2}x}{{\left\{\begin{array}{c}(si{n}^{2}x+co{s}^{2}x)(si{n}^{3}x+co{s}^{3}x)\end{array}\right\}}^2}dx$

$\Rightarrow \int \frac{si{n}^{2}xco{s}^{2}x}{{\left\{\begin{array}{c}(si{n}^{3}x+co{s}^{3}x)\end{array}\right\}}^2}dx$

$\Rightarrow \int \frac{si{n}^{2}xco{s}^{2}x}{\left\{\begin{array}{c}si{n}^{6}x+2si{n}^{3}xco{s}^{3}x+co{s}^{6}x\end{array}\right\}}dx$

on dividing numerator and denominator by $co{s}^{6}x$we get

$\Rightarrow \int \frac{ta{n}^{2}xse{c}^{2}x}{\left\{\begin{array}{c}ta{n}^{6}x+2ta{n}^{3}x+1\end{array}\right\}}dx$

$\Rightarrow \int \frac{ta{n}^{2}xse{c}^{2}x}{{\left(\begin{array}{c}1+ta{n}^{3}x\end{array}\right)}^2}dx$

Let $(1+ta{n}^{3}x)=t\Rightarrow 3ta{n}^{2}xse{c}^{2}xdx=dt$

$\Rightarrow \frac{1}{3}\int \frac{dt}{{\left(\begin{array}{c}t\end{array}\right)}^2}=-\frac{1}{3}\frac{1}{t}+C$

$=\frac{-1}{3(1+ta{n}^{3}x)}+C$

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