Mathematics

Evaluate:
$$\int sin^4 x cos^4x dx $$


SOLUTION
$$I=\displaystyle \int \sin^4x. \cos^4x.dx$$

We know that

$$\sin^2x=\dfrac {1-\cos 2x}{2},\ \cos^2x=\dfrac {1+\cos 2x}{2}$$

$$\displaystyle \int \left (\dfrac {1-\cos 2x}{2}\right)^2. \left (\dfrac {1+\cos 2x}{2}\right)^2dx$$

$$=\dfrac {1}{16}\displaystyle \int (1-\cos^22x)^2dx$$

$$=\dfrac {1}{16}\displaystyle \int (1+\cos^42x-2\cos^2 2x)dx$$

$$=\dfrac {1}{16}\displaystyle \int 1+\dfrac {(1+\cos 4x)^2}{4}-2\left (\dfrac {1+\cos 4x}{2}\right)dx$$

$$=\dfrac {1}{16}\displaystyle \int 1+\dfrac {1+\cos^24x+2\cos 4x}{2}-1-\cos 4x \ dx$$

$$=\dfrac {1}{16}\displaystyle \int \dfrac {1+\dfrac {1+\cos 8x}{2}+2\cos 4x-2\cos 4x}{2}dx$$

$$=\dfrac {1}{32}\displaystyle \int 1+\dfrac {1+\cos 8x}{2}dx$$

$$=\dfrac {1}{64}\displaystyle \int 3+\cos 8x\ dx$$

$$=\dfrac {1}{64} \left [3x+\dfrac {\sin 8x}{8}\right]+c$$

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