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#### Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 60 maths.

Answer :   $\frac{1}{6}$

Hint: Use indefinite integral formula and the limits to solve this integral

Given: $\int_{0}^{\frac{\pi}{4}}\frac{\sin^2x.\cos ^2x}{(\sin ^3x+\cos ^3x)}dx$

Solution: $\int_{0}^{\frac{\pi}{4}}\frac{\sin^2x.\cos ^2x}{(\sin ^3x+\cos ^3x)}dx$

Dividing numerator and denominator by $\cos ^6x$

$\int_{0}^{\frac{\pi}{4}} \frac{\frac{\sin ^{2} x \cdot \cos ^{2} x}{\cos ^{6} x}}{\frac{\left(\sin ^{3} x+\cos ^{3} x\right)^{2}}{\cos ^{6} x}} d x=\int_{0}^{\frac{\pi}{4}} \frac{\frac{\sin ^{2} x}{\cos ^{2} x} \cdot \frac{\cos ^{2} x}{\cos ^{4} x}}{\left(\frac{\sin ^{3} x+\cos ^{3} x}{\cos ^{3} x}\right)^{2}} d x=\int_{0}^{\frac{\pi}{4}} \frac{\tan ^{2} x \frac{1}{\cos ^{2} x}}{\left(\frac{\sin ^{3} x+\cos ^{3} x}{\cos ^{3} x}\right)^{2}} d x$

\begin{aligned} &=\int_{0}^{\frac{\pi}{4}} \frac{\tan ^{2} x \sec ^{2} x}{\left(\frac{\sin ^{3} x}{\cos ^{3} x}+\frac{\cos ^{3} x}{\cos ^{3} x}\right)^{2}} d x \\ &=\int_{0}^{\frac{\pi}{4}} \frac{\tan ^{2} x \sec ^{2} x}{\left(\tan ^{3} x+1\right)^{2}} d x=\int_{0}^{\frac{\pi}{4}} \frac{\tan ^{2} x \sec ^{2} x}{\tan ^{6} x+1+2 \tan ^{3} x} d x \end{aligned}
Put $\tan x =t \Rightarrow \sec^2x\; dx=dt$
Now x=0  then t=0 & when $x=\frac{\pi}{4}$ then t=1
\begin{aligned} &\therefore \int_{0}^{\frac{\pi}{4}} \frac{\sin ^{2} x \cdot \cos ^{2} x}{\left(\sin ^{3} x+\cos ^{3} x\right)^{2}} d x=\int_{0}^{\frac{\pi}{4}} \frac{\tan ^{2} x \sec ^{2} x}{\tan ^{6} x+1+2 \tan ^{3} x} d x \\ &=\int_{0}^{1} \frac{t^{2}}{1+t^{6}+2 t^{3}} d t \\ &=\int_{0}^{1} \frac{t^{2}}{1+\left(t^{3}\right)^{2}+2 t^{3}} d t \\ &=\int_{0}^{1} \frac{t^{2}}{\left(1+t^{3}\right)^{2}} d t \end{aligned}

Again putting $1+\frac{t}{3}=u\Rightarrow 3t^2dt=du\Rightarrow t^2dt=\frac{du}{3}$
then
\begin{aligned} &=\int_{0}^{1} \frac{t^{2}}{\left(1+t^{3}\right)^{2}} d t=\int_{0}^{1} \frac{1}{u^{2}} \frac{d u}{3}=\frac{1}{3} \int_{0}^{1} \frac{1}{u^{2}} d u \\ &=\frac{1}{3} \int_{0}^{1} u^{-2} d u=\frac{1}{3}\left[\frac{u^{-2+1}}{-2+1}\right]_{0}^{1} \\ &=\frac{1}{3}\left[\frac{u^{-1}}{-1}\right]_{0}^{1}=-\frac{1}{3}\left[\frac{1}{u}\right]_{0}^{1} \\ &=-\frac{1}{3}\left[\frac{1}{1+t^{3}}\right]_{0}^{1}=-\frac{1}{3}\left[\frac{1}{1+1}-\frac{1}{1+0}\right] \\ &=-\frac{1}{3}\left[\frac{1}{2}-1\right]=-\frac{1}{3}\left[\frac{1-2}{2}\right] \\ &=\frac{1}{6} \end{aligned}