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

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

Answers (1)

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}

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