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#### Need Solution for R D  Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 11 Maths Textbook Solution.

Answer: $x^{2} \sin x+2 x \cos x-2 \sin x+c$

Hint: Use integration by parts

$\int u v d x=u \int v d x-\int \frac{d}{d x} u \int v d x$

Given: Let $I=\int x^{2} \cos x d x$

Solution: $I=\int x^{2} \cos x d x$

\begin{aligned} &=x^{2} \int \cos x d x-\int\left(\frac{d}{d x} x^{2} \int \cos x d x\right) d x \\ &=x^{2} \sin x-2\left[\int x \sin x d x\right] \end{aligned}

Now apply by part method again

\begin{aligned} &=x^{2} \sin x-2 \int x \sin x d x \\ &=x^{2} \sin x-2\left[x \int \sin x d x-\int\left(\frac{d}{d x} x \cdot \int \sin x d x\right) d x\right] \\ &=x^{2} \sin x-2\left[x(-\cos x)-\int 1 .(-\cos x) d x\right] \\ &=x^{2} \sin x-2(-x \cos x+\sin x)+c \end{aligned}

On further simplification,

$=x^{2} \sin x+2 x \cos x-2 \sin x+c$

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