Question

Integral $\int \frac{{\sin }^{2}x{\cos }^{2}x}{({\sin }^{5}x+{\cos }^{3}x{\sin }^{2}x+{\sin }^{3}x{\cos }^{2}x+{\cos }^{5}x{)}^2}dx$ is equal to :

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

Answer 1

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

$\int \frac{si{n}^{2}xco{s}^{2}x}{(si{n}^{5}x+co{s}^{3}xsi{n}^{2}x+si{n}^{3}xco{s}^{2}x+co{s}^{5}x{)}^2}$

$=\int \frac{si{n}^{3}xco{s}^{2}x}{\left[si{n}^{3}x\right(si{n}^{2}x+co{s}^{2}x)+co{s}^{3}x(si{n}^{2}x+co{s}^{2}x){]}^2}$

$=\int \frac{si{n}^{2}xco{s}^{2}x}{[si{n}^{3}x+co{s}^{3}x{]}^2}dx$

$=\int \frac{si{n}^{2}xco{s}^{2}x}{co{s}^{6}x[ta{n}^{3}x+1{]}^2}$

$=\int \frac{ta{n}^{2}xse{c}^{2}x}{[ta{n}^{3}x+1{]}^2}dx$

Let $t=tanx\Rightarrow dt=se{c}^{2}xdx$

$=\int \frac{t^2}{(t^3+1{)}^2}dt=\frac{1}{3}\int \frac{3t^2}{(t^3+1{)}^2}dt$

Let $t^3+1=h$

$\Rightarrow 3t^{2}dt=dh$

$=\frac{1}{3}\int \frac{dk}{k^2}=\frac{1}{3}\left(\frac{-1}{h}\right)+C$

$=\frac{1}{3}\left(\frac{-1}{t^3+1}\right)+C$

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