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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 : 62 of 271
Marks: +1, -0
sin⁡2(2x+5)\sin^2 (2x + 5)
Solution:  
Let I=∫sin⁡2(2x+5d)xI=\int \sin^2(2x+5d)x =12[1−cos⁡2(2x+5)]dx=\frac{1}{2}\left[1-\cos 2(2x+5)\right]dx
=12∫[1−cos⁡(4x+10)]dx=\frac{1}{2}\int \left[1-\cos(4x+10)\right]dx =12[x−sin⁡(4x+10)4]+C=\frac{1}{2}\left[x-\frac{\sin(4x+10)}{4}\right]+C
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