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Evaluate the following problems using second fundamental theorem: $\int\limits_{0}^{\pi/2}\sin^{2}xdx$

$\begin{array}{1 1} \frac{\pi}{4} \\ \frac{\pi}{6} \\ \frac{\pi}{2} \\ \pi \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 $\int \limits_0^{\large\frac{\pi}{2}} \sin ^ 2 x dx$
Step 1:
$\int \limits_0^{\large\frac{\pi}{2}} \sin ^ 2 x dx= \int \limits_0^{\large\frac{\pi}{2}} \large\frac{1-\cos 2x}{2}$$dx$
Step 2:
$\qquad=\large\frac{1}{2}$$x-\large\frac{1}{4} $$\sin 2x\bigg]_0^{\large\frac{\pi}{2}}$
$\qquad=\large\frac{\pi}{4}$
answered Aug 13, 2013 by meena.p
 

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