Question

$\int {{{\sin }^2}(\ln x)dx} $ is equal to
A) $\dfrac{{2x}}{{15}}(5 + 2\sin (2\ln x)) + \cos (2\ln x)) + c$
B) $\dfrac{x}{{10}}(5 + 2\sin (2\ln x)) - \cos (2\ln x)) + c$
C) $\dfrac{x}{{10}}(5 - 2\sin (2\ln x)) - \cos (2\ln x)) + c$
D) $\dfrac{x}{{10}}(5 - 2\sin (2\ln x)) + \sin (2\ln x)) + c$

Answer 1

Hint: To solve this, we will start with putting, ${\sin ^2}(\ln x) = \dfrac{{1 - \cos (2\ln x)}}{2}$, then separating the constant and trigonometric value, we will get I $ = \dfrac{1}{2}[\int {dx - \int {\cos (2\ln x)dx]} } $, Now we will solve the I1 I1$ = \int {\cos (2\ln x)dx} $. On assuming $\ln x = t$, we will get, I1 $ = \int {\cos (2\ln x)dx} $in the form I1 $ = \int {{e^t}\cos 2tdt} $. Afterwards using necessary conditions, we will further solve the integration to get the required value.

Complete step-by-step answer:
 We have been given $\int {{{\sin }^2}(\ln x)dx} $
Now, we know that, $\cos 2\theta = 1 - 2{\sin ^{\mathbf{2}}}\theta $
$\begin{gathered}
  2{\sin ^{\mathbf{2}}}\theta = 1 - \cos 2\theta \\
  {\sin ^{\mathbf{2}}}\theta = \dfrac{{1 - \cos 2\theta }}{2} \\
\end{gathered} $
Now, on putting the value of ${\sin ^2}(\ln x) = \dfrac{{1 - \cos (2\ln x)}}{2}$ in the given expression, we get $\int {\dfrac{{1 - \cos (2\ln x)}}{2}dx} $
Let, I $ = \dfrac{1}{2}[\int {dx - \int {\cos (2\ln x)dx]} } $
Here, we will assume, I1 $ = \int {\cos (2\ln x)dx} $, and we will solve it separately.
Now, let $\ln x = t$
$ \Rightarrow x = {e^t}$
On differentiating both sides, we get
$dx = {e^t}dt$
Now putting the above value in I1 $ = \int {\cos (2\ln x)dx} $, we get
I1 $ = \int {{e^t}\cos 2tdt} $
We know that, $\int {{e^{ax}}\cos bx = \dfrac{{{e^{ax}}}}{{{a^2} + {b^2}}}[a\cos bx + b\sin bx]} $
So, on comparing I1 $ = \int {{e^t}\cos 2tdt} $ with above expression, we get a = 1 and b = 2, so now on substituting the values, we get
$\begin{gathered}
   \Rightarrow \dfrac{{{e^t}}}{{{1^2} + {2^2}}}[\cos 2t + 2\sin 2t] \\
   \Rightarrow \dfrac{{{e^t}}}{5}[\cos 2t + 2\sin 2t] \\
\end{gathered} $
Since, we have taken, t = lnx, then
I1 $ = \dfrac{{{e^{\ln x}}}}{5}[\cos (2\ln x) + 2\sin 2\ln x]$
Now, ${e^{\ln x}} = x$
So we get, I1 $ = \dfrac{x}{5}[\cos (2\ln x) + 2\sin 2\ln x]$
Now putting I1 $ = \dfrac{x}{5}[\cos (2\ln x) + 2\sin 2\ln x]$ in I $ = \int {\dfrac{1}{2}dx - \int {\cos (2\ln x)dx} } $, we get
I $ = \int {\dfrac{1}{2}dx - \dfrac{1}{2}[\dfrac{x}{5}\{ \cos (2\ln x) + 2\sin 2\ln x\} ]} $
I $ = \dfrac{x}{2} - \dfrac{x}{{10}}[\cos (2\ln x) + 2\sin 2\ln x] + c$
I $ = \dfrac{x}{{10}}[5 - \cos (2\ln x) - 2\sin 2\ln x] + c$
Thus, option (C) $\dfrac{x}{{10}}(5 - 2\sin (2\ln x)) - \cos (2\ln x)) + c$, is correct.
So, the correct answer is “Option C”.

Note: To solve this question, we have taken few values, like ${\sin ^{\mathbf{2}}}\theta = \dfrac{{1 - \cos 2\theta }}{2}$, students should take care of changing the angles, as we can’t use $\angle \theta $, we have to use $\ln x,$because that’s how the question is given. And students should also take care that here we have taken $\ln x = t$, but the question is given in terms of x, so we need to get the answers in terms of x only.