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