#### Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 42 Maths Textbook Solution.

Answer:   $I= \frac{\pi}{5}$

Hint: In this equation we have  $\int_{0}^{a} f(x)$   formula

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

Solution:

\begin{aligned} &\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x \\ & \end{aligned}

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

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

\begin{aligned} &=\pi \int_{0}^{\pi} \sin x \cdot \cos ^{4} x d x-\int_{0}^{\pi} x \cdot \sin x \cdot \cos ^{4} x d x \\ & \end{aligned}

$I=\frac{\pi}{2} \int_{0}^{\pi} \sin x \cdot \cos ^{4} x d x-I$

$I=-\frac{\pi}{2} \int_{1}^{-1} t^{4} d t$                                                           $\left[\begin{array}{l} \cos x=t \\ \sin x d x=d t \\ x \rightarrow 0 \\ t \cos x=1 \end{array}\right]$

\begin{aligned} &I=\frac{\pi}{2} \int_{-1}^{1} t^{4} d t \\ & \end{aligned}

$I=\frac{\pi}{10}\left[t^{5}\right]_{-1}^{1} \\$

$I=\frac{\pi}{10}\left[1^{5}-(-1)^{5}\right] \\$

$I=\frac{\pi}{10}(1+1) \\$

$I=\frac{2 \pi}{10} \\$

$I=\frac{\pi}{5}$

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