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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.3
Find the integrals of the functions in Exercises 62 to 83.
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Question : 67 of 271
Marks: +1, -0
sin⁡xsin⁡2xsin⁡3x\sin x \sin 2x \sin 3x
Solution:  
Let I=∫sin⁡xsin⁡2xsin⁡3xdxI=\int \sin x \sin 2x \sin 3x dx
=12∫(2sin⁡xsin⁡2x)sin⁡3xdx=\frac{1}{2}\int (2\sin x \sin 2x) \sin 3x dx
=12∫(cos⁡x−cos⁡2x)sin⁡3xdx    [∵2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)]=\frac{1}{2}\int (\cos x - \cos 2x) \sin 3x dx\;\;[\because 2\sin A \sin B = \cos(A-B) - \cos(A+B)]
=14∫2sin⁡3xcos⁡xdx−14∫2sin⁡3xcos⁡3xdx    [∵2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)]=\frac{1}{4}\int 2\sin 3x \cos x dx - \frac{1}{4}\int 2\sin 3x \cos 3x dx\;\;[\because 2\sin A \cos B = \sin(A+B) + \sin(A-B)]
=14∫(sin⁡4x+sin⁡2x)dx−14∫sin⁡6xdx=\frac{1}{4}\int (\sin 4x + \sin 2x) dx - \frac{1}{4}\int \sin 6x dx
=−116cos⁡4x−18cos⁡2x+124cos⁡6x+C=-\frac{1}{16}\cos 4x - \frac{1}{8}\cos 2x + \frac{1}{24}\cos 6x + C
=14[16cos⁡6x−14cos⁡4x−12cos⁡2x]+C=\frac{1}{4}\left[\frac{1}{6}\cos 6x - \frac{1}{4}\cos 4x - \frac{1}{2}\cos 2x\right] + C
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