#### Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 61 maths.

Answer :   $\frac{8}{21}$

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

Given: $\int_{0}^{\frac{\pi}{2}}\sqrt{\cos x-\cos^2x}(\sec^2 x-1)\cos^2xdx$

Solution:

$\int_{0}^{\frac{\pi}{2}}\sqrt{\cos x-\cos^2x}(\sec^2 x-1)\cos^2xdx$
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \sqrt{\cos x\left(1-\cos ^{2} x\right)}\left(\sec ^{2} x-1\right) \cos ^{2} x d x \\ &=\int_{0}^{\frac{\pi}{2}} \sqrt{\cos x \sin ^{2} x}\left(\tan ^{2} x\right) \cos ^{2} x d x \mid \\ &=\int_{0}^{\frac{\pi}{2}} \sqrt{\cos x} \sin x\left(\frac{\sin ^{2} x}{\cos ^{2} x}\right) \cos ^{2} x d x \\ &=\int_{0}^{\frac{\pi}{2}} \sqrt{\cos x} \sin ^{3} x d x \end{aligned}

Put $\cos x=t\Rightarrow -\sin xdx=dt\Rightarrow \sin xdx=-dt$
Now x=0 then t=1 & when $x=\frac{\pi}{2}$  then t=0
\begin{aligned} &\therefore \int_{0}^{\frac{\pi}{2}} \sqrt{\cos x-\cos ^{3} x}\left(\sec ^{2} x-1\right) \cos ^{2} x d x=\int_{0}^{\frac{\pi}{2}} \sqrt{\cos x} \sin ^{3} x d x \\ &=\int_{1}^{0} \sqrt{t} \sin ^{2} x(-d t) \\ &=-\int_{1}^{0} \sqrt{t}\left(1-\cos ^{2} x\right) d t \end{aligned}

\begin{aligned} &=-\int_{1}^0 \sqrt{t}\left(1-t^{2}\right) d t=-\int_{1}^0\left(\sqrt{t}-t^{2} \sqrt{t}\right) d t \\ &=-\int_{1}^{0}\left(t^{\frac{1}{2}}-t^{2+\frac{1}{2}}\right) d t=-\int_{1}^{0}\left(t^{\frac{1}{2}}-t^{\frac{5}{2}}\right) d t \\ &=-\int_{1}^{0} t^{\frac{1}{2}} d t+\int_{1}^{0} t^{\frac{5}{2}} d t \\ &=-\left[\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{1}^{0}+\left[\frac{t^{\frac{5}{2}+1}}{\frac{5}{2}+1}\right]_{1}^{0} \end{aligned}

\begin{aligned} &=-\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{0}+\left[\frac{t^{\frac{7}{2}}}{7}\right]_{1}^{0} \\ &=\frac{-2}{3}\left[o^{\frac{3}{2}}-1^{\frac{3}{2}}\right]+\frac{2}{7}\left[o^{\frac{7}{2}}-1^{\frac{7}{2}}\right] \\ &=\frac{-2}{3}[0-1]+\frac{2}{7}[0-1] \\ &=\frac{2}{3}-\frac{2}{7} \\ &=\frac{14-6}{21}=\frac{8}{21} \end{aligned}