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Evaluate the following problems using second fundamental theorem: $\int\limits_{0}^{\large\frac{\pi}{4}}2\sin^{2}x \sin 2x dx$

$\begin{array}{1 1} \frac{1}{4} \\ \frac{3}{4} \\ \frac{5}{4} \\ \frac{7}{4} \end{array} $

1 Answer

Toolbox:
  • If $F(x)=\int \limits_a^x f(t)dt $ then $\int \limits_a^b f(x) dx=F(b)-F(a)$
Given $\int \limits_0^{\large\frac{\pi}{4}} 2 \sin ^2 x \sin 2x dx$
Step 1:
$\int \limits_0^{\large\frac{\pi}{4}} 2 \sin ^2 x \sin 2x dx=2\int \limits_0^{\large\frac{\pi}{4}} \sin ^2 x .2 \sin x \cos x dx $
Step 2:
$\qquad=4\int \limits_0^{\large\frac{\pi}{4}} \sin ^3 x \cos x dx $
$\qquad=\large\frac{4 \sin ^4 x}{4} \bigg]_0^{\large\frac{\pi}{4}}$
$\qquad=\sin ^4 \large\frac{\pi}{4}$$-\sin ^4 0$
$\qquad=\large\frac{1}{4}$
answered Aug 13, 2013 by meena.p
 

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