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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 : 65 of 271
Marks: +1, -0
sin⁡3(2x+1)\sin^3 (2x + 1)
Solution:  
Let I=∫sin⁡3(2x+1)dxI=\int \sin^3 (2x + 1)dx
=14∫3sin⁡(2x+1)−sin⁡3(2x+1)dx=\frac{1}{4}\int 3\sin(2x + 1)-\sin 3(2x + 1)dx [∵sin⁡3θ=3sin⁡θ−4sin⁡3θ][\because \sin 3\theta = 3 \sin\theta - 4 \sin 3\theta]
=34(−cos⁡(2x+1)2)−14(−cos⁡3(2x+1)6)+C=\frac{3}{4}\left(-\frac{\cos(2x+1)}{2}\right)-\frac{1}{4}\left(-\frac{\cos 3(2x+1)}{6}\right)+C
=−38cos⁡(2x+1)+124cos⁡3(2x+1)+C=-\frac{3}{8}\cos(2x+1)+\frac{1}{24}\cos 3(2x+1)+C
=−38cos⁡(2x+1)+124[4cos⁡3(2x+1)−3cos⁡(2x+1)]+C    [∵cos⁡3θ=4cos⁡3θ−3cos⁡θ]=-\frac{3}{8}\cos(2x+1)+\frac{1}{24}\left[4\cos^3(2x+1)-3\cos(2x+1)\right]+C\;\; [\because \cos^3\theta = 4\cos^3\theta - 3\cos\theta]
=−38cos⁡(2x+1)+16cos⁡3(2x+1)−18cos⁡(2x+1)+C=-\frac{3}{8}\cos(2x+1)+\frac{1}{6}\cos^3(2x+1)-\frac{1}{8}\cos(2x+1)+C
=−12cos⁡(2x+1)+16cos⁡3(2x+1)+C=-\frac{1}{2}\cos(2x+1)+\frac{1}{6}\cos^3(2x+1)+C
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