#### Provide solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 12 textbook solution.

Answer:- $\frac{\pi}{5}$

Hints:-  You must know the integration rules of trigonometric functions and its limits

Given:-  $\int_{0}^{\pi} \sin x \cos ^{4} x \cdot d x$

Solution : $\int_{0}^{\pi} \sin x \cos ^{4} x \cdot d x=I$                                            ......(1)

$I=\int_{0}^{\pi}(\pi-x) \sin (\pi-x) \cos ^{4}(\pi-x) \cdot d x\\$

\begin{aligned} &I=\int_{0}^{\pi}(\pi-x) \sin x \cos ^{4} x \cdot d x\\ \end{aligned}                                                     $\text { .......(2) }[\because \sin (\pi-x)=\sin x]$

$\mathrm{Eq}(1)+\mathrm{Eg}(2) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\cos (\pi-x)=-\cos x]$

\begin{aligned} &2 I=\int_{0}^{\pi} x \sin x \cdot \cos ^{4} x d x+\int_{0}^{\pi}(\pi-x) \sin x \cdot \cos ^{4} x \cdot d x \\ &2 I=\int_{0}^{\pi} \pi \sin x \cdot \cos ^{4} x \cdot d x \end{aligned}

Let $\sin x = t$

\begin{aligned} &x=0, x=\pi \\ &t=0, t=0 \\ &\cos x d x=d t \\ &2 I=\int_{0}^{0} \text { No benefit. } \end{aligned}

Now, Let $\cos x = t$

\begin{aligned} &x=0, t=1 \\ &x=\pi, t=-1 \end{aligned}

\begin{aligned} &2 I=\int_{1}^{-1} \pi t^{4} d t \\ &I=\frac{-\pi}{2} \int_{1}^{-1} t^{4} d t \\ &I=\frac{-\pi}{2}\left[\frac{t^{5}}{5}\right]_{1}^{-1} \end{aligned}

\begin{aligned} &=\frac{-\pi}{2}\left[\frac{-1}{5}-\frac{1}{5}\right] \\ &=\frac{-\pi}{2}\left[\frac{-2}{5}\right] \\ &=\frac{\pi}{5} \end{aligned}