#### Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 19

Answer:   $\frac{2}{35}-\frac{9}{280 \sqrt{2}}$

Given:   $\int_{0}^{\frac{\pi}{4}} \cos ^{4} x \sin ^{3} x d x$

Hint: Use trigonometric identities.

Solution:   $\int_{0}^{\frac{\pi}{4}} \cos ^{4} x \sin ^{3} x d x$

\begin{aligned} &=\int_{0}^{\frac{\pi}{4}} \cos ^{4} x \sin x\left(1-\cos ^{2} x\right) d x \\ & \end{aligned}

$\left(\begin{array}{l} \because \sin ^{2} x+\cos ^{2} x=1 \\ \sin ^{2} x=1-\cos ^{2} x \end{array}\right)$

\begin{aligned} &\text { Let } \\ & \end{aligned}

$\cos x=t \\$                          (Differentiate w.r.t to x)

$-\sin x d x=d t$

\begin{aligned} &=-\int_{0}^{\frac{\pi}{4}} t^{4}\left(1-t^{2}\right) d t \\ & \end{aligned}

$=\int_{0}^{\frac{\pi}{4}} t^{4}\left(t^{2}-1\right) d t \\$

$=\int_{0}^{\frac{\pi}{4}}\left(t^{6}-t^{4}\right) d t=\left(\frac{t^{7}}{7}-\frac{t^{5}}{5}\right)_{0}^{\frac{\pi}{4}}$

\begin{aligned} &=\left(\frac{\cos ^{7} x}{7}-\frac{\cos ^{5} x}{5}\right)_{0}^{\frac{\pi}{4}}=\frac{1}{(\sqrt{2})^{7}} \times \frac{1}{7}-\frac{1}{(\sqrt{2})^{5}} \times \frac{1}{5}-\frac{1}{7}+\frac{1}{5} \\ & \end{aligned}

$=\frac{1}{56 \sqrt{2}}-\frac{1}{20 \sqrt{2}}+\frac{2}{35}$

\begin{aligned} &=\frac{1}{4 \sqrt{2}}\left(\frac{1}{14}-\frac{1}{5}\right)+\frac{2}{35} \\ & \end{aligned}

$=\frac{2}{35}+\frac{1}{4 \sqrt{2}}\left(\frac{5-14}{70}\right) \\$

$=\frac{2}{35}-\frac{9}{280 \sqrt{2}}$