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Q10  Find the integrals of the functions \sin ^ 4 x

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\sin ^ 4 x  can be written as follows using trigonometric identities
\\=\sin^2x.\sin^2x\\ =\frac{1}{4}(1-\cos 2x)^{2}\\ =\frac{1}{4}(1+\cos^22x-2\cos 2x)\\ =\frac{1}{4}(1+\frac{1}{2}(1+\cos 4x)-2\cos 2x)\\ =3/8+\frac{\cos 4x}{8}-\frac{\cos 2x}{2}
 

Therefore, 
\Rightarrow \int \sin^4x\ dx = \int \frac{3}{8}dx+\frac{1}{8}\int\cos 4x\ dx -\frac{1}{2}\int\cos 2x\ dx
                                 = \frac{3x}{8}+\frac{\sin 4x}{32} -\frac{\sin 2x}{4}+C

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manish

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