Question

integrate this question wrt x

Answer 1

hello friend

》》》》》see the above attachment《《《《

☆☆☆☆hope it hlps

Answer 2

We have to find $\int {\sin^4 x \cos ^4x} \, dx$. Now

$\sin x \cos x=\frac{1}{2} \sin (2x)\\ \sin^2 x \cos^2 x=\frac{1}{4} \sin^2 (2x)\\ \sin^2 x \cos^2 x=\frac{1}{8}[ 1-\cos (4x)]\\ \sin^4 x \cos^4 x=\frac{1}{64}[ 1-\cos (4x)]^2\\ \sin^4 x \cos^4 x=\frac{1}{64}[ 1-2\cos (4x)+\cos ^2(4x)]$

$\sin^4 x \cos^4 x=\frac{1}{128}[ 2-4\cos (4x)+2\cos ^2(4x)]\\ \sin^4 x \cos^4 x=\frac{1}{128}[ 2-4\cos (4x)+1+\cos (8x)]\\ \sin^4 x \cos^4 x=\frac{1}{128}[ 3-4\cos (4x)+\cos (8x)]$

Thus the integral is

$\int {\sin^4 x \cos ^4x} \, dx= \int \frac{1}{128}[ 3-4\cos (4x)+\cos (8x)]\,dx\\ \int {\sin^4 x \cos ^4x} \, dx= \frac{3x}{128}-\frac{1}{128}\sin (4x)+\frac{1}{1024}\sin (8x)+C$