Question

integrate this question
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Answer 1

hope it will be help ful for u..

Answer 2

Topic:

Integration

Solution:

We need to evaluate the following integral.

$\displaystyle\int_0^{\pi/2} \sin^2(x)\ dx$

To solve this integral, we use the following formula,

$\boxed{\sin^2(x) = \dfrac{1- \cos(2x)}{2}}$

Using this, we get:
$\displaystyle\longrightarrow\int_0^{\pi/2} \dfrac{1-\cos(2x)}{2}\ dx$

$\displaystyle\longrightarrow\int_0^{\pi/2}\dfrac12 - \dfrac{\cos(2x)}{2}\ dx$

$\displaystyle\longrightarrow\int_0^{\pi/2}\dfrac12 \, dx -\int_0^{\pi/2} \dfrac{\cos(2x)}{2}\ dx$

Using the following identities:
$\boxed{\int x^n\ dx = \dfrac{x^{n+1}}{n+1} + C}$

$\boxed{\int\cos(ax) \ dx = \dfrac{\sin(ax)}{a } +C}$

We get the following result:

$\displaystyle\longrightarrow\dfrac12 \Big(x\Big)_0^{\pi/2} -\bigg( \dfrac{\sin(2x)}{2\times 2}\bigg)_0^{\pi/2}$

$\displaystyle\longrightarrow\dfrac12 \Big(\dfrac{\pi}{2} - 0\Big) -\bigg( \dfrac{\sin(\pi)}{4} - \dfrac{\sin(0)}{4}\bigg)$

$\displaystyle\longrightarrow\dfrac\pi4-(0-0)$

$\displaystyle\longrightarrow\dfrac\pi4$

This is the required answer.