Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 57 Maths Textbook Solution.

Answer: $\frac{x\sin 3x}{12}+\frac{\cos 3x}{36}+\frac{3x\sin x}{4}+\frac{3\cos x}{4}+c$

Hint: $\cos ^{3} x=\frac{\cos 3x+3\cos x}{4}$

Given:$\int x\cos ^{3}xdx$

Solution:

$\therefore \cos ^{3}x=\frac{\cos 3x+3\cos x}{4}$

$\therefore \int x \cos ^{3} x d x=\int x\left[\frac{\cos 3 x+3 \cos x}{4}\right] d x=\frac{1}{4} \int x \cos 3 x d x+\frac{3}{4} \int x \cos x d x$

\begin{aligned} &\Rightarrow \frac{1}{4}\left[x \int \cos 3 x d x-\int\left[\frac{d x}{d x}\left(\int \cos 3 x d x\right)\right] d x\right]+\frac{3}{4}\left[x \int \cos x d x-\int\left[\frac{d x}{d x} \int \cos x d x\right] d x\right] \\ &\Rightarrow \frac{1}{4}\left[\frac{x \sin 3 x}{3}-\int \frac{\sin 3 x}{3} d x\right]+\frac{3}{4}\left(x \sin x-\int \sin x d x\right) \\ &\Rightarrow \frac{1}{4}\left[\frac{x \sin 3 x}{3}+\frac{\cos 3 x}{9}\right]+\frac{3}{4}(x \sin x+\cos x)+c \\ &=\frac{x \sin 3 x}{12}+\frac{\cos 3 x}{36}+\frac{3 x \sin x}{4}+\frac{3 \cos x}{4}+c \end{aligned}