Question

Solve the following integral

$\int_{0}^{\pi} (sin2x) dx$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int_{0}^{\pi}\rm sin2x \: dx$

Let assume that

$\rm :\longmapsto\:I \: = \: \displaystyle\int_{0}^{\pi}\rm sin2x \: dx - - - (1)$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{0}^{a}\rm f(x)dx = \displaystyle\int_{0}^{a}f(a - x) \: dx}}}$

So, using this identity, we get

$\rm :\longmapsto\:I \: = \: \displaystyle\int_{0}^{\pi}\rm sin2(\pi - x)\: dx$

$\rm :\longmapsto\:I \: = \: \displaystyle\int_{0}^{\pi}\rm sin(2\pi - 2x)\: dx$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ sin(2\pi - x) = - sinx \: }}}$

So, using this, we get

$\rm :\longmapsto\:I \: = \: - \: \displaystyle\int_{0}^{\pi}\rm sin2x \: dx$

$\rm :\longmapsto\:I \: = \: - \: I$

$\rm :\longmapsto\:2I \: = \: 0$

$\rm\implies \:I = 0$

$\rm\implies \:\:\displaystyle\int_{0}^{\pi}\rm sin2x \: dx \: = \: 0$

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MORE TO KNOW

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{a}^{b}\rm f(x)dx = \displaystyle\int_{a}^{b}f(y) \: dy}}}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{a}^{b}\rm f(x)dx = \: - \: \displaystyle\int_{b}^{a}f(x) \: dx}}}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{a}^{b}\rm f(x)dx = \displaystyle\int_{a}^{b}f(a + b - x) \: dx}}}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{ - a}^{a}\rm f(x)dx = 0 \: if \: f( - x) = - f(x)}}}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{ - a}^{a}\rm f(x)dx =2\displaystyle\int_{0}^{a} \rm f(x) \: dx\: if \: f( - x) = f(x)}}}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{0}^{2a}\rm f(x)dx = 0 \: if \: f(2a - x) = - f(x)}}}$

Answer 2

Question:-

$\sf \: Solve \: the \: following \: integral \\ \sf \: \int_{0}^{\pi} (sin2x) dx</p><p>$

Solution:-

$\sf \large \int_{0}^{\pi} (sin2x) dx$

$\sf \large \longmapsto \int_{0}^{\pi} (sin2x) dx$

$\sf \large \longmapsto \bigg [ - \frac { 1}{2} cos2x \bigg] ^\pi_0$

$\sf \large \longmapsto \bigg[ - \frac{1}{2} \: cos2\pi \bigg] - \bigg[ - \frac{1}{2} \: cos \: 2 \times 0 \bigg]$

$\sf \large \longmapsto \bigg[ - \frac{1}{2} \times 1 \bigg] - \bigg[ - \frac{1}{2} \times 1 \bigg]$

$\sf \large \longmapsto - \frac{1}{2} - ( - \frac{1}{2} )$

$\sf \large \longmapsto{{ \cancel{ - \frac{1}{2}}}}{{ \cancel{ + \frac{1}{2} }}} = 0.$

Hence, Answer is 0.

Answer:-

$\sf \large { \boxed{ \blue{\int_{0}^{\pi} (sin2x) dx=0}}}$

Hope you have satisfied. ⚘