#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.25 Question 16 Maths Textbook Solution.

Answer: $\frac{x^{3}}{6}-\frac{x^{2} \sin 2 x}{4}-\frac{x \cos 2 x}{4}+\frac{\sin 2 x}{8}+c$

Hint: Use the formula

$1) \sin ^{2} x=\frac{1-\cos 2 x}{2}$

$\text { 2) } \int u v d x=u \int v d x-\int \frac{d}{d x} u \int v d x$

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

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

\begin{aligned} &\int x^{2}\left(\frac{1-\cos 2 x}{2}\right) d x=\int \frac{x^{2}}{2}-\frac{x^{2} \cos 2 x}{2} d x \\ &=\int \frac{x^{2}}{2} d x-\int \frac{x^{2} \cos 2 x}{2} d x \\ &=\frac{x^{3}}{3 \times 2}-\frac{1}{2} \int x^{2} \cos 2 x d x \\ &=\frac{x^{3}}{6}-\frac{1}{2}\left(x^{2} \int \cos 2 x d x-\int\left(\frac{d}{d x} x^{2} \int \cos 2 x d x\right) d x\right) \\ &=\frac{x^{3}}{6}-\frac{1}{2}\left(x^{2} \frac{\sin 2 x}{2}-\int 2 x \frac{\sin 2 x}{2} d x\right) \\ &=\frac{x^{3}}{6}-\frac{1}{2}\left(x^{2} \frac{\sin 2 x}{2}-\int x \sin 2 x d x\right) \end{aligned}

Now by integrating the second part we have,

\begin{aligned} &=\frac{x^{3}}{6}-\frac{1}{2}\left(x^{2} \frac{\sin 2 x}{2}-\left[x \frac{-\cos 2 x}{2}-\int 1 \frac{-\cos 2 x}{2} d x\right]\right) \\ &=\frac{x^{3}}{6}-\frac{1}{2}\left(x^{2} \frac{\sin 2 x}{2}+\frac{x \cos 2 x}{2}-\frac{\sin 2 x}{4}\right)+c \\ &=\frac{x^{3}}{6}-\frac{x^{2} \sin 2 x}{4}-\frac{x \cos 2 x}{4}+\frac{\sin 2 x}{8}+c \end{aligned}