The integral \int \sin ^2x\cos ^2x \over ( \sin ^5x + \cos ^3x\sin ^2x + \sin ^3x\cos ^2x + \cos ^5x )^2 dx is equal to

- 1 \over 3( 1 + \tan ^3x ) + C

Indefinite Integrals PYQ — JEE Main 2018 · Shift 0 · Q39 with Solution

View written solutionFree

Correct answer: C

Simplify the denominator

We need to evaluate

$I=\int \frac{\sin^2 x\cos^2 x}{\left(\sin^5 x+\cos^3 x\sin^2 x+\sin^3 x\cos^2 x+\cos^5 x\right)^2}\,dx.$

Let

$D=\sin^5 x+\cos^3 x\sin^2 x+\sin^3 x\cos^2 x+\cos^5 x.$

Group terms:

$D=\sin^2 x(\sin^3 x+\cos^3 x)+\cos^2 x(\sin^3 x+\cos^3 x).$

So,

$D=(\sin^2 x+\cos^2 x)(\sin^3 x+\cos^3 x)=\sin^3 x+\cos^3 x.$

Hence the integral becomes

$I=\int \frac{\sin^2 x\cos^2 x}{(\sin^3 x+\cos^3 x)^2}\,dx.$

Rewrite using $\tan x$

Divide numerator and denominator by $\cos^6 x$ :

$I=\int \frac{\tan^2 x\sec^{-2}x}{(1+\tan^3 x)^2}\,dx$

but it is cleaner to directly substitute using

$\sin x=\tan x\cos x.$

Then

$\sin^2 x\cos^2 x=\tan^2 x\cos^4 x,$

and

$\sin^3 x+\cos^3 x=\cos^3 x(\tan^3 x+1).$

Therefore,

=\frac{\tan^2 x\cos^4 x}{\cos^6 x(1+\tan^3 x)^2} =\frac{\tan^2 x\sec^2 x}{(1+\tan^3 x)^2}.$$ So, $$I=\int \frac{\tan^2 x\sec^2 x}{(1+\tan^3 x)^2}\,dx.$$ --- 3. **Substitute** Let $$t=\tan x \implies dt=\sec^2 x\,dx.$$ Then $$I=\int \frac{t^2}{(1+t^3)^2}\,dt.$$ Now let $$u=1+t^3 \implies du=3t^2\,dt,$$ so $$t^2\,dt=\frac{du}{3}.$$ Thus, $$I=\frac13\int u^{-2}\,du =\frac13\left(-u^{-1}\right)+C =-\frac{1}{3u}+C.$$ Substituting back, $$I=-\frac{1}{3(1+t^3)}+C=-\frac{1}{3(1+\tan^3 x)}+C.$$ --- 4. **Match with options** This is exactly $$\boxed{-\frac{1}{3(1+\tan^3 x)}+C}.$$ So the correct option is **C**. --- 5. **Verification with stored answer** Stored correct answer: **C**. Our derived answer also gives **C**, so they agree.

More from Indefinite Integrals

View written solutionFree

Correct answer: C

Simplify the denominator

We need to evaluate

$I=\int \frac{\sin^2 x\cos^2 x}{\left(\sin^5 x+\cos^3 x\sin^2 x+\sin^3 x\cos^2 x+\cos^5 x\right)^2}\,dx.$

Let

$D=\sin^5 x+\cos^3 x\sin^2 x+\sin^3 x\cos^2 x+\cos^5 x.$

Group terms:

$D=\sin^2 x(\sin^3 x+\cos^3 x)+\cos^2 x(\sin^3 x+\cos^3 x).$

So,

$D=(\sin^2 x+\cos^2 x)(\sin^3 x+\cos^3 x)=\sin^3 x+\cos^3 x.$

Hence the integral becomes

$I=\int \frac{\sin^2 x\cos^2 x}{(\sin^3 x+\cos^3 x)^2}\,dx.$

Rewrite using $\tan x$

Divide numerator and denominator by $\cos^6 x$ :

$I=\int \frac{\tan^2 x\sec^{-2}x}{(1+\tan^3 x)^2}\,dx$

but it is cleaner to directly substitute using

$\sin x=\tan x\cos x.$

Then

$\sin^2 x\cos^2 x=\tan^2 x\cos^4 x,$

and

$\sin^3 x+\cos^3 x=\cos^3 x(\tan^3 x+1).$

Therefore,

=\frac{\tan^2 x\cos^4 x}{\cos^6 x(1+\tan^3 x)^2} =\frac{\tan^2 x\sec^2 x}{(1+\tan^3 x)^2}.$$ So, $$I=\int \frac{\tan^2 x\sec^2 x}{(1+\tan^3 x)^2}\,dx.$$ --- 3. **Substitute** Let $$t=\tan x \implies dt=\sec^2 x\,dx.$$ Then $$I=\int \frac{t^2}{(1+t^3)^2}\,dt.$$ Now let $$u=1+t^3 \implies du=3t^2\,dt,$$ so $$t^2\,dt=\frac{du}{3}.$$ Thus, $$I=\frac13\int u^{-2}\,du =\frac13\left(-u^{-1}\right)+C =-\frac{1}{3u}+C.$$ Substituting back, $$I=-\frac{1}{3(1+t^3)}+C=-\frac{1}{3(1+\tan^3 x)}+C.$$ --- 4. **Match with options** This is exactly $$\boxed{-\frac{1}{3(1+\tan^3 x)}+C}.$$ So the correct option is **C**. --- 5. **Verification with stored answer** Stored correct answer: **C**. Our derived answer also gives **C**, so they agree.

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