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.