Question

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

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

Answer 1

The correct option is D

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

Explanation for the correct option:

$$\begin{gathered} \int \frac{{\sin }^{2}x{\cos }^{2}x}{{({\sin }^{5}x+{\cos }^{3}x{\sin }^{2}x+{\sin }^{3}x{\cos }^{2}x+{\cos }^{5}x)}^2}dx \\ =\int \frac{{\sin }^{2}x{\cos }^{2}x}{{\left\{{\sin }^{2}x\left({\sin }^{3}x+{\cos }^{3}x\right)+{\cos }^{2}x\left({\sin }^{3}x+{\cos }^{3}x\right)\right\}}^2}dx \\ =\int \frac{{\sin }^{2}x{\cos }^{2}x}{{\left\{\left({\sin }^{2}x+{\cos }^{2}x\right)\left({\sin }^{3}x+{\cos }^{3}x\right)\right\}}^2}dx \\ =\int \frac{{\sin }^{2}x{\cos }^{2}x}{{\left\{\left({\sin }^{3}x+{\cos }^{3}x\right)\right\}}^2}dx \end{gathered}$$

Divide by ${\cos }^{3}x$ in numerator and denominator we get

$=\int \frac{se{c}^{2}x·{\tan }^{2}x}{{({\tan }^{3}x+1)}^2}dx$

Let $1+{\tan }^{3}x=t$

$3{\tan }^{2}xse{c}^{2}xdx=dt$

$$\begin{gathered} =\frac{1}{3}\int \frac{dt}{t^2}=-\frac{1}{3}\frac{1}{t}+C \\ =-\frac{1}{3\left(1+{\tan }^{3}x\right)}+C \end{gathered}$$

Hence the correct option is option(d) i.e. $-\frac{1}{3(1+{\tan }^{3}x)}+C$.