Question

Evaluate the following integral
\[\int{{{e}^{\log \left( \sin x \right)}}dx=}\]
(a) sin x + c
(b) – cos x + c
(c) \[{{e}^{\log \cos x}}+c\]
(d) None of these

Answer 1

Hint: Let’s consider the given integral. Now, by using \[{{a}^{\log b}}={{b}^{\log a}},\text{ convert }{{e}^{\log \left( \sin x \right)}}\text{ into }{{\left( \sin x \right)}^{\log e}}\]. Now we know that the value of log e = 1, so substituting this value will result in the simplified form of the given function as sin x. Then, we can solve the integral further.

Complete step-by-step answer:

In this question, we have to evaluate the integral
\[\int{{{e}^{\log \left( \sin x \right)}}dx}\]
Let us consider the integral given in the question.
\[I=\int{{{e}^{\log \left( \sin x \right)}}dx}\]
We know that \[{{a}^{\log b}}={{b}^{\log a}}\]. By substituting a = e and b = sin x, we get,
\[{{e}^{\log \left( \sin x \right)}}={{\left( \sin x \right)}^{\log e}}\]
By substituting the value of \[{{e}^{\log \left( \sin x \right)}}\] in the above integral, we get,
\[I=\int{{{\left( \sin x \right)}^{\log e}}dx}\]
We know that log e = 1. By substituting the value of log e, we get,
\[I=\int{\left( \sin x \right)dx}\]
We know that \[\int{\sin xdx=-\cos x+c}\]. By using this in the above integral, we get,
\[I=-\cos x+c\]
So, we get,
\[\int{{{e}^{\log \left( \sin x \right)}}dx=-\cos x+c}\]
Hence, option (b) is the right answer.

Note: In this question, many students do not solve the expression inside the integral first, that is they do not convert \[{{e}^{\log \left( \sin x \right)}}\text{ into }\left( \sin x \right)\] which makes the question seem very tough. So, they must properly examine the expression before finding its integral. Also, in this question, many students make this mistake of writing \[\int{\sin xdx}\] as cos x + c but actually, it is – cos x + c. So, this must be taken care of. In this question, students can also cross-check their answer by differentiating – cos x + c and checking if it is equal to sin x or not as follows:
\[D=\dfrac{d}{dx}\left( -\cos +c \right)\]
We know that,
\[\dfrac{d}{dx}\cos x=-\sin x\text{ and }\dfrac{d}{dx}\left( \text{constant} \right)=0\]
So, we get,
D = – (– sin x) + 0 = sin x which is equal to the initial expression. So, our answer is correct.