Browse Questions

# Integrate : $\int \sin ^2 x \cos ^2 x \;dx$

$\begin {array} {1 1} (a)\;\frac{x}{8}-\frac{\sin 4x}{32}+c \\ (b)\;\frac{x}{8}-\frac{\sin 4x}{8}+c \\ (c)\;\frac{x}{8}+\frac{\sin x}{16}+c \\ (d)\;None \end {array}$

$\int \large\frac{1}{cosec^2 x. \sec ^2 x }$$dx By trignometry property => \int \large\frac{cosec^2 x- \cot ^2 x}{cosec ^2 x \; sec ^2 x }$$dx$
=> $\int (\cos^2 x)- \large\frac{\cos ^2 x. \sin ^2 x . \cos ^2 x }{\sin ^2 x }.$$dx => \int (\cos ^ 2 x - \cos ^4 x)dx => \int \bigg(\large\frac{1+\cos 2 x}{2}\bigg)$$dx -\int \cos ^4 x$
=> $\int \bigg(\large\frac{1+\cos 2 x}{2}\bigg)$$dx -\int (1+\cos 2x)^2 dx => \int \bigg(\large\frac{1+\cos 2 x}{2}\bigg)$$dx -\large\frac{1}{4}$$[(1+ \cos ^2 2x)+ 2 \cos 2x] => \int \large\frac{1}{2} +\frac{\cos 2 x}{2}-\frac{1}{4}-\frac{\cos ^2 2x}{4} -\frac{1}{2}$$\cos 2x $$dx -\int \bigg(\large\frac{1}{4} - \frac{\cos ^2 2x}{4} \bigg)$$dx$
=> $\int \large\frac{1}{4} $$(1- \cos^2 2x).dx => \int \large\frac{1}{4}$$\{1-(\frac{1}{2} (1+ \cos 4x))dx$