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Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.25 Question 7 Maths Textbook Solution.

Answers (1)

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

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

                   =x^{2} \int \cos x d x-\int \frac{d}{d x} x^{2} \int(\cos x d x) d x

By integrating w.r.t x

                   =x^{2} \sin x-\int 2 x \sin x d x

So we get,

                =x^{2} \sin x-2\left[\int x \sin x d x\right]

Now apply by parts method

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