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Q1  Find the integrals of the functions \sin ^ 2 ( 2x+ 5 )

Answers (1)

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\sin ^ 2 ( 2x+ 5 )

using the trigonometric identity

 sin^2x=\frac{1-cos2x}{2} 

we can write the given question as 

= \frac{1-\cos 2(2x+5)}{2}= \frac{1-\cos (4x+10)}{2}
\\=\int \frac{1-\cos (4x+10)}{2}dx\\ =\frac{1}{2}\int dx - \frac{1}{2}\int \cos(4x+10)dx\\ =\frac{x}{2}-\frac{1}{2}[\sin(4x+10)/4]\\ =\frac{x}{2}-\frac{\sin(4x+10)}{8}+C

 

                    

Posted by

manish

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