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$I=\int \limits_0^{\frac{\pi}{2}} \sin ^4 x \cos^2 x dx$ find the value of I

$\begin {array} {1 1} (a)\;\large\frac{\pi}{16} \\ (b)\;\large\frac{\pi}{8} \\ (c)\;\large\frac{\pi}{32} \\ (d)\;0 \end {array}$
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$I= \int \limits_0^{\frac{\pi}{2}} \sin ^4 x \cos ^2 x dx$-------(1)
Using $ \int \limits_a^b f(x) dx =\int \limits _0^b f(a+b-x) dx$
$I= \int \limits _0^{\frac{\pi}{2}} \cos ^4 x \sin ^2 x dx $-----------(2)
adding (1) and (2)
$2I= \int \limits _0^{\frac{\pi}{2}} \sin ^2 x \cos ^2 x (\sin ^2 x+ \cos ^2 x)dx $-----------(2)
$\qquad = \int \limits _0^{\frac{\pi}{2}} \large\frac{1}{4}$$ \sin ^ 2 2x dx $
$\qquad= \int \limits _0^{\frac{\pi}{2}} \large\frac{1}{4} \frac{(1- \cos 4x)}{2}$$dx$
$\qquad= \large\frac{1}{8} \bigg[x - \frac{\sin 4x}{4}\bigg]_0^{\frac{\pi}{2}}$
$\qquad= \large\frac{1}{8}\big(\large\frac{\pi}{2}$$-0\big)$
$\qquad= \large\frac{\pi}{16}$
answered Dec 20, 2013 by meena.p
edited Mar 20, 2014 by rvidyagovindarajan_1
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