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

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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


\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} \\


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