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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 21 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 21 Maths Textbook Solution.

Answers (1)

Answer:

I=\frac{1}{4}\left(x+\frac{\sin 6 x}{6}+\frac{\sin 4 x}{4}+\frac{\sin 2 x}{2}\right)+c

Given:

\int \cos x \cos 2 x \cos 3 x d x

Hint:

To solve this equation  we will use trigonometry method.

Solution: 

\int(\cos x \cos 2 x) \cos 3 x d x

I=\int \frac{\cos (x+2 x)+\cos (x-2 x)}{2} \cos 3 x d x \quad\left[\because \cos C \cdot \cos D=\frac{1}{2} \cos (C+D)+\cos (C-D)\right)

I=\frac{1}{2} \int(\cos 3 x+\cos x) \cos 3 x d x

I=\frac{1}{2}\left(\int \cos ^{2} 3 x d x+\int \cos x \cos 3 x\right) d x

I=\frac{1}{2} \int \frac{1+\cos 6 x}{2} d x+\frac{1}{2} \int \cos (x+3 x)+\cos (x-2 x) d x

I=\frac{1}{2} \int 1+\frac{\cos 6 x}{2} d x+\frac{1}{2} \int(\cos 4 x+\cos (2 x)) d x

I=\frac{1}{2} \int \frac{1}{2} d x+\int \frac{\cos 6 x}{2} d x+\frac{1}{2} \int \cos 4 x d x+\frac{1}{2} \int \cos 2 x d x

I=\frac{1}{2}\left(\frac{1}{2} x+\frac{1}{2} \frac{\sin 6 x}{6}+\frac{1}{2} \frac{\sin 4 x}{4}+\frac{1}{2} \frac{\sin 2 x}{2}\right)+c

I=\frac{1}{2}\left(\frac{x}{2}+\frac{\sin 6 x}{12}+\frac{\sin 4 x}{8}+\frac{\sin 2 x}{4}\right)+c

 

Posted by

infoexpert21

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