#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 6 Maths Textbook Solution.

Answer: $\frac{x^{2} \sin 2 x}{2}+\frac{x \cos 2 x}{2}-\frac{\sin 2 x}{4}+c$

Hint:Use integration by parts

\begin{aligned} &\int u v d x=u \int v d x-\int \frac{d}{d x} u \int v d x \\ &\text { Let } u=x^{2} \& v=\cos 2 x \end{aligned}

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

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

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

Integrate on by parts applying once again

\begin{aligned} &=\frac{x^{2} \sin 2 x}{2}-\left[x\left(\frac{-\cos 2 x}{2}\right)-\int\left(\frac{-\cos 2 x}{2}\right) d x\right] \\ &=\frac{x^{2} \sin 2 x}{2}-\left[\frac{-x \cos 2 x}{2}+\int \frac{\cos 2 x}{2} d x\right] \\ &=\frac{x^{2} \sin 2 x}{2}-\left[\frac{-x \cos 2 x}{2}+\frac{\sin 2 x}{4}\right]+c \\ &=\frac{x^{2} \sin 2 x}{2}+\frac{x \cos 2 x}{2}-\frac{\sin 2 x}{4}+c \end{aligned}