Question

Find the value of following integral:
 $\int_0^\pi {\sqrt {1 + \sin 2x} dx} $.

Answer 1

Hint – In order to solve this problem you need to use the formulas of ${(\cos x + \sin x)^2}$ and sin 2x. Then solve further to get the right answer to this problem.

Complete step-by-step answer:
The given equation is $\int_0^\pi {\sqrt {1 + \sin 2x} dx} $.
We can write $1 + \sin 2x = {\cos ^2}x + {\sin ^2} + 2\sin x\cos x\,\,\,\,\,(\because {\cos ^2}x + {\sin ^2} = 1,\, \sin2x = 2\sin x\cos x\,\,)$
and ${\cos ^2}x + {\sin ^2} + 2\sin x\cos x$ is nothing but ${(\cos x + \sin x)^2}$ as ${(a + b)^2} = {a^2} + {b^2} + 2ab$.
So, $\int_0^\pi {\sqrt {1 + \sin 2x} dx} $ can be written as $\int_0^\pi {\sqrt {{{(\cos x + \sin x)}^2}} dx} = \int_0^\pi {(\cos x + \sin x)dx} $.
So, on integrating now we get,
$
   \Rightarrow \int_0^\pi {(\cos x + \sin x)dx = \int_0^\pi {\cos xdx + } \int_0^\pi {\sin x} dx} \\
   \Rightarrow \sin x\left| {_0^\pi - \cos x\left| {_0^\pi } \right.} \right. \\
   \Rightarrow \sin \pi - \sin 0 - (\cos \pi - \cos 0) \\
   \Rightarrow 0 - 0 - ( - 1 - 1) \\
   \Rightarrow 1 + 1 = 2 \\
$
Hence, the answer to this question is 2.

Note – In this question we need to know the formula of ${(\cos x + \sin x)^2}$ and should also know that integration of sin is –cos and cos is sin. Then put the integration limit and get the answer to the question by solving algebraically.