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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 20 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 20  Maths Textbook Solution.

Answers (1)

Answer:

-\frac{\cos 4 x}{16}-\frac{\cos 2 x}{8}+\frac{\cos 6 x}{24}+c

Given:

\int \sin x \sin 2 x \sin 3 x d x

Hint:

Use the formula 2 \sin A \sin B=-\cos (A+B)+\cos (A-B)\rfloor

Solution: 

\int \sin x \sin 2 x \sin 3 x d x

=\int \frac{1}{2}[\cos (x-2 x)-\cos (x+2 x)] \sin 3 x d x

=\int \frac{1}{2}(\cos (-\mathrm{x})-\cos 3 \mathrm{x}) \sin 3 \mathrm{x} \mathrm{dx}

=\frac{1}{2} \int \cos x \sin 3 x-\cos 3 x \sin 3 x d x

=\frac{1}{2} \int \frac{1}{2}(\sin 4 x+\sin 2 x) d x-\frac{1}{2} \int \frac{1}{2}(\sin 6 x) d x

=\frac{1}{4}\left[\int(\sin 4 x+\sin 2 x-\sin 6 \mathrm{x}) \mathrm{dx}\right]

=\frac{1}{4}\left[-\frac{\cos 4 x}{4}-\frac{\cos 2 x}{2}+\frac{\cos 6 x}{6}\right]+c

-\frac{\cos 4 x}{16}-\frac{\cos 2 x}{8}+\frac{\cos 6 \mathrm{x}}{24}+\mathrm{c}

 

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