Question

find the value of integration 0_^π/3 √1-cos2x/2 dx​

Answer 1

Step-by-step explanation:

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Answer 2

Given :-

$\rm :\longmapsto\:\displaystyle\int^{\dfrac{\pi}{3} } _{0} \sqrt{\dfrac{1 - cos2x}{2} } dx$

Identities Used :-

$\boxed{ \bf \: 1 - cos2x = 2 {sin}^{2}x}$

$\boxed{ \bf \:\:\displaystyle\int \: \bf \: sinx dx = - cosx + c}$

Solution :-

$\rm :\longmapsto\:\displaystyle\int^{\dfrac{\pi}{3} } _{0} \sqrt{\dfrac{1 - cos2x}{2} } dx$

$\rm \: \: = \: \: \:\displaystyle\int^{\dfrac{\pi}{3} } _{0} \sqrt{\dfrac{2 {sin}^{2}x }{2} } dx$

$\rm \: \: = \: \: \:\displaystyle\int^{\dfrac{\pi}{3} } _{0} \sqrt{ {sin}^{2}x} dx$

$\rm \: \: = \: \: \:\displaystyle\int^{\dfrac{\pi}{3} } _{0} |sinx| dx$

$\: \: \: \: \: \: \: \: \: \: \: \: \because \: \boxed{ \bf \: {( \sqrt{x}) }^{2} = |x| }$

$\rm \: \: = \: \: \:\displaystyle\int^{\dfrac{\pi}{3} } _{0}sinx dx$

$\red{ \because\boxed{ \bf \: As \: x \in \: {1}^{st} \: quadrant \: \implies \: sinx > 0}}$

$\rm \: \: = \: - \: \bigg(cosx\bigg) ^{\dfrac{\pi}{3} } _{0}$

$\rm \: \: = \: \: - \: \bigg(cos\dfrac{\pi}{3} - cos0\bigg)$

$\rm \: \: = \: \: - \: ( \dfrac{1}{2} - 1)$

$\rm \: \:=\: \: \dfrac{1}{2}$

$\rm :\implies\:\displaystyle\int^{\dfrac{\pi}{3} } _{0} \bf \: \sqrt{\dfrac{1 - cos2x}{2} } dx = \dfrac{1}{2}$

Additional Information

$\sf (1). \: \int\limits^b_a {f(x)} \, dx = \int\limits^b_a {f(t)}$

$\sf (2). \: \int\limits^b_a {f(x)} \, dx = -\int\limits^a_b {f(x)}$

$\sf (3). \: \int\limits^b_a {f(x)} \, dx = \int\limits^c_a {f(x)} \, dx + \int\limits^b_c {f(x)} \, dx \ where\ \ a < c < b$

$\sf (4). \: \int\limits^a_0 {f(x)} \, dx =\int\limits^a_0 {f(a - x)}$

$\sf (5). \: \int\limits^b_a {f(x)} \, dx = \int\limits^b_a{f(a + b - x)}$