Question

The integral $\int \frac{si{n}^{2}xco{s}^{2}x}{(si{n}^{5}x+co{s}^{3}xsi{n}^{2}x+si{n}^{3}xco{s}^{2}x+co{s}^{5}x{)}^2}dx$ is equal to

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

Answer 1

The correct option is A $\frac{-1}{3(1+ta{n}^{3}x)}+C$

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

$=\int \frac{{\sin }^{2}x{\cos }^{2}x}{\left[{\cos }^{6}x{(\frac{{\sin }^{5}x}{{\cos }^{3}x}+\frac{{\sin }^{3}x}{\cos x}+{\sin }^{2}x+{\cos }^{2}x)}^2\right]}dx$

$=\int \frac{\frac{{\sin }^{2}x}{{\cos }^{2}x}.\frac{1}{{\cos }^{2}x}}{\left[{(\frac{{\sin }^{3}x}{{\cos }^{3}x}.{\sin }^{2}x+{\sin }^{2}x.\frac{\sin x}{\cos x}+{\sin }^{2}x+{\cos }^{2}x)}^2\right]}dx$

$=\int \frac{{\sec }^{2}x.{\tan }^{2}x}{{({\tan }^{3}x.{\sin }^{2}x+{\sin }^{2}x.\tan x+1)}^2}dx$ $[\because \frac{1}{\sec x}=\cos x,{\sin }^{2}x+{\cos }^{2}x=1]$

$=\int \frac{{\sec }^{2}x.{\tan }^{2}x}{{(\tan x.{\sin }^{2}x+(1+{\tan }^{2}x)+1)}^2}dx$

$=\int \frac{{\sec }^{2}x.{\tan }^{2}x}{{(\tan x.{\sin }^{2}x+{\sec }^{2}x+1)}^2}dx$ $[\because 1+{\tan }^{2}x={\sec }^{2}x]$

$=\int \frac{{\sec }^{2}x.{\tan }^{2}x}{{(1+\tan x.\frac{{\sin }^{2}x}{{\cos }^{2}x})}^2}dx$

$=\int \frac{{\sec }^{2}x.{\tan }^{2}x}{{(1+{\tan }^{3}x)}^2}$

$=\frac{1}{3}\int \frac{d(1+{\tan }^{3}x)}{{(1+{\tan }^{3}x)}^2}$

$=\frac{1}{3}\frac{{(1+{\tan }^{3}x)}^{-2+1}}{-2+1}$

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

Hence, the answer is $\frac{-1}{3(1+{\tan }^{3}x)}+c.$