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Question

Question: Solve the following integral: \[\int{\sqrt{1+\sin 2xdx}}=\] (a) sin x + cos x + c (b) sin x – ...

Solve the following integral:
1+sin2xdx=\int{\sqrt{1+\sin 2xdx}}=
(a) sin x + cos x + c
(b) sin x – cos x + c
(c) cos x – sin x + c
(d) None of these

Explanation

Solution

Hint: In order to find the solution of this question, we need to remember a few trigonometric identities like sin2x+cos2x=1,sin2x=2sinxcosx{{\sin }^{2}}x+{{\cos }^{2}}x=1,\sin 2x=2\sin x\cos x and a few more algebraic identities like (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab and then a few standard integrations like sinxdx=cosx+c\int{\sin xdx=-\cos x+c} and cosxdx=sinx+c.\int{\cos xdx=\sin x+c}. By using these formulas, we can find the correct answer to this question.

Complete step-by-step answer:
In this question, we have been asked to integrate the function 1+sin2x\sqrt{1+\sin 2x} with respect to x. So, to solve this question, we will first consider 1+sin2xdx\int{\sqrt{1+\sin 2x}}dx as I. So, we can write,
I=1+sin2xdxI=\int{\sqrt{1+\sin 2x}}dx
Now, we know that, sin2x+cos2x=1.{{\sin }^{2}}x+{{\cos }^{2}}x=1. So, we get,
I=sin2x+cos2x+sin2xdxI=\int{\sqrt{{{\sin }^{2}}x+{{\cos }^{2}}x+\sin 2x}dx}
Now, we know that 2 sin x cos x = sin 2x. So, we can get,
I=sin2x+cos2x+2sinxcosxdxI=\int{\sqrt{{{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x}dx}
Now, we know that, (a+b)2=a2+b2+2ab.{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab. So, for a = sin x and b = cos x, we get,
sin2x+cos2x+2sinxcosx=(sinx+cosx)2{{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x={{\left( \sin x+\cos x \right)}^{2}}
Therefore, we can write I as,
I=(sinx+cosx)2dxI=\int{\sqrt{{{\left( \sin x+\cos x \right)}^{2}}}dx}
Now, we know that, a2=a.\sqrt{{{a}^{2}}}=a. So, we can write I as,
I=(sinx+cosx)dxI=\int{\left( \sin x+\cos x \right)dx}
And we know that it can be further written as,
I=sinxdx+cosxdxI=\int{\sin xdx}+\int{\cos x}dx
We also know that, sinxdx=cosx+c1 and cosxdx=sinx+c2.\int{\sin xdx}=-\cos x+{{c}_{1}}\text{ and }\int{\cos xdx}=\sin x+{{c}_{2}}. Therefore, we can write I as,
I=cosx+sinx+c1+c2I=-\cos x+\sin x+{{c}_{1}}+{{c}_{2}}
I=sinxcosx+cI=\sin x-\cos x+c
Hence, we can say, 1+sin2xdx=sinxcosx+c.\int{\sqrt{1+\sin 2x}dx=\sin x-\cos x+c}.
Therefore, we get the option (b) as the right answer.

Note: While solving the question, one can think of differentiating each option one by one and then see which matches the question but it could be complicated as well as a lengthy method and we can also make calculation mistakes by this method. So, it is better to use the conventional method and get the right answer.