Question

integration of √1-cos2x​

Answer 1

Answer :-

$\implies \boxed{ \sf{ \sqrt{2} ( - \cos x) + \: c \: (constant)}} \:$

Explanation :-

$\implies \sf{ \int \: \sqrt{1 - \cos2x} \: dx \: }$

We know that ;

$\leadsto \boxed{ \: \sf{ \cos2 \theta = 1 - 2 {\sin}^{2} \theta \: }} \\ \\ \therefore \: \\ \\ \implies \sf{ \int \: \sqrt{1 - (1 - 2 { \sin}^{2}x) } \: dx} \\ \\ \implies \int \sf{ \sqrt{1 - 1 + 2 { \sin}^{2}x } \: dx }\: \\ \\ \implies \sf{ \int \: \sqrt{2} \: \times \sqrt{ { \sin}^{2} x} \: dx \: } \\ \\ \implies \sf{ \sqrt{2} \int \: \sin x \: \: dx \: }$

$\because \boxed{ \sf{ \int \sin x \: \: dx = - \cos \: x \: + c \: }} \\ \\ \therefore \: \: \\ \implies \sf{ \sqrt{2} \int \sin x \: \: dx \: } \\ \\ \implies \boxed{ \sf{ \sqrt{2} ( - \cos x) + \: c \: (constant)}}$

Answer 2

AnsWer :

$\tt -\sqrt{2} \cos x + c.$

Solution :

We have,

$\implies\tt \int \sqrt{1 - \cos 2x \: dx }$

$\therefore \: \tt1 - \cos \: 2x = 2 {\sin }^{2} x$

$\implies \tt\int \sqrt{2 { \sin \: }^{2}x \: dx }$

$\implies \tt\int \sqrt{2} . \sqrt{ { \sin }^{2} x \: dx}$

$\therefore \: \tt here \: \sqrt{2} \: is \: constant , \: take \: common.$

$\implies \tt \sqrt{2} \int \sqrt{ { \sin}^{2} x \: dx}$

$\implies \tt \sqrt{2} \int \sin x \: dx$

$\implies \tt \sqrt{2} ( - \cos x) + c$

$\implies \tt - \sqrt{2} \cos x + c.$

Therefore, the value of integration √ 1- cos 2x dx is -√2 cos x +c.

$\rule{200}3$

$\boxed{\begin{minipage}{7 cm} Identities \\ \\ \cos2x = \cos^2x-\sin^2x\\ \\ \cos2x = 1-2\sin^2x \\ \\ \cos2x = 2\cos^2x-1 \\ \\ 1-\cos2x = 2\sin^2x \end{minipage}}$