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=
(a) sin x + cos x + c
(b) sin x – cos x + c
(c) cos x – sin x + c
(d) None of these
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 and a few more algebraic identities like (a+b)2=a2+b2+2ab and then a few standard integrations like ∫sinxdx=−cosx+c and ∫cosxdx=sinx+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 with respect to x. So, to solve this question, we will first consider ∫1+sin2xdx as I. So, we can write,
I=∫1+sin2xdx
Now, we know that, sin2x+cos2x=1. So, we get,
I=∫sin2x+cos2x+sin2xdx
Now, we know that 2 sin x cos x = sin 2x. So, we can get,
I=∫sin2x+cos2x+2sinxcosxdx
Now, we know that, (a+b)2=a2+b2+2ab. So, for a = sin x and b = cos x, we get,
sin2x+cos2x+2sinxcosx=(sinx+cosx)2
Therefore, we can write I as,
I=∫(sinx+cosx)2dx
Now, we know that, a2=a. So, we can write I as,
I=∫(sinx+cosx)dx
And we know that it can be further written as,
I=∫sinxdx+∫cosxdx
We also know that, ∫sinxdx=−cosx+c1 and ∫cosxdx=sinx+c2. Therefore, we can write I as,
I=−cosx+sinx+c1+c2
I=sinx−cosx+c
Hence, we can say, ∫1+sin2xdx=sinx−cosx+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.