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Q 4 Find the integrals of the functions \sin ^ 3 ( 2x +1 )

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\int \sin^3(2x+1)dx = \int \sin^2(2x+1).\sin(2x+1)dx

The integral can be written as 

                                     = \int (1-\cos^2(2x+1)).\sin(2x+1)dx
Let 
\\\cos (2x+1) =t\\ \sin (2x+1)dx = -dt/2

                                        \\=\frac{-1}{2}\int (1-t^2)dt\\ =\frac{-1}{2}[t-t^3/3]\\ =\frac{t^3}{6}-\frac{t}{2}

Now,  replace the value of t, we get;

=\frac{\cos^3(2x+1)}{6}-\frac{\cos(2x+1)}{2}+C

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manish

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