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Question: \(\int {{{\sin }^2}(\ln x)dx} \) is equal to A) \(\dfrac{{2x}}{{15}}(5 + 2\sin (2\ln x)) + \cos (2...

∫sin⁡2(ln⁡x)dx\int {{{\sin }^2}(\ln x)dx} is equal to
A) 2x15(5+2sin⁡(2ln⁡x))+cos⁡(2ln⁡x))+c\dfrac{{2x}}{{15}}(5 + 2\sin (2\ln x)) + \cos (2\ln x)) + c
B) x10(5+2sin⁡(2ln⁡x))−cos⁡(2ln⁡x))+c\dfrac{x}{{10}}(5 + 2\sin (2\ln x)) - \cos (2\ln x)) + c
C) x10(5−2sin⁡(2ln⁡x))−cos⁡(2ln⁡x))+c\dfrac{x}{{10}}(5 - 2\sin (2\ln x)) - \cos (2\ln x)) + c
D) x10(5−2sin⁡(2ln⁡x))+sin⁡(2ln⁡x))+c\dfrac{x}{{10}}(5 - 2\sin (2\ln x)) + \sin (2\ln x)) + c

Explanation

Solution

Hint : To solve this, we will start with putting, sin⁡2(ln⁡x)=1−cos⁡(2ln⁡x)2{\sin ^2}(\ln x) = \dfrac{{1 - \cos (2\ln x)}}{2}, then separating the constant and trigonometric value, we will get I =12[∫dx−∫cos⁡(2ln⁡x)dx]= \dfrac{1}{2}[\int {dx - \int {\cos (2\ln x)dx]} }, Now we will solve the I1 I1=∫cos⁡(2ln⁡x)dx= \int {\cos (2\ln x)dx}. On assuming ln⁡x=t\ln x = t, we will get, I1 =∫cos⁡(2ln⁡x)dx= \int {\cos (2\ln x)dx}in the form I1 =∫etcos⁡2tdt= \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 ∫sin⁡2(ln⁡x)dx\int {{{\sin }^2}(\ln x)dx}
Now, we know that, cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2{\sin ^{\mathbf{2}}}\theta
2sin⁡2θ=1−cos⁡2θ sin⁡2θ=1−cos⁡2θ2 \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)=1−cos⁡(2ln⁡x)2{\sin ^2}(\ln x) = \dfrac{{1 - \cos (2\ln x)}}{2} in the given expression, we get ∫1−cos⁡(2ln⁡x)2dx\int {\dfrac{{1 - \cos (2\ln x)}}{2}dx}
Let, I =12[∫dx−∫cos⁡(2ln⁡x)dx]= \dfrac{1}{2}[\int {dx - \int {\cos (2\ln x)dx]} }
Here, we will assume, I1 =∫cos⁡(2ln⁡x)dx= \int {\cos (2\ln x)dx}, and we will solve it separately.
Now, let ln⁡x=t\ln x = t
⇒x=et\Rightarrow x = {e^t}
On differentiating both sides, we get
dx=etdtdx = {e^t}dt
Now putting the above value in I1 =∫cos⁡(2ln⁡x)dx= \int {\cos (2\ln x)dx}, we get
I1 =∫etcos⁡2tdt= \int {{e^t}\cos 2tdt}
We know that, ∫eaxcos⁡bx=eaxa2+b2[acos⁡bx+bsin⁡bx]\int {{e^{ax}}\cos bx = \dfrac{{{e^{ax}}}}{{{a^2} + {b^2}}}[a\cos bx + b\sin bx]}
So, on comparing I1 =∫etcos⁡2tdt= \int {{e^t}\cos 2tdt} with above expression, we get a = 1 and b = 2, so now on substituting the values, we get
⇒et12+22[cos⁡2t+2sin⁡2t] ⇒et5[cos⁡2t+2sin⁡2t] \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 =eln⁡x5[cos⁡(2ln⁡x)+2sin⁡2ln⁡x] = \dfrac{{{e^{\ln x}}}}{5}[\cos (2\ln x) + 2\sin 2\ln x]
Now, eln⁡x=x{e^{\ln x}} = x
So we get, I1 =x5[cos⁡(2ln⁡x)+2sin⁡2ln⁡x] = \dfrac{x}{5}[\cos (2\ln x) + 2\sin 2\ln x]
Now putting I1 =x5[cos⁡(2ln⁡x)+2sin⁡2ln⁡x] = \dfrac{x}{5}[\cos (2\ln x) + 2\sin 2\ln x] in I =∫12dx−∫cos⁡(2ln⁡x)dx= \int {\dfrac{1}{2}dx - \int {\cos (2\ln x)dx} }, we get
I =∫12dx−12[x5cos⁡(2ln⁡x)+2sin⁡2ln⁡x]= \int {\dfrac{1}{2}dx - \dfrac{1}{2}[\dfrac{x}{5}\\{ \cos (2\ln x) + 2\sin 2\ln x\\} ]}
I =x2−x10[cos⁡(2ln⁡x)+2sin⁡2ln⁡x]+c = \dfrac{x}{2} - \dfrac{x}{{10}}[\cos (2\ln x) + 2\sin 2\ln x] + c
I =x10[5−cos⁡(2ln⁡x)−2sin⁡2ln⁡x]+c = \dfrac{x}{{10}}[5 - \cos (2\ln x) - 2\sin 2\ln x] + c
Thus, option (C) x10(5−2sin⁡(2ln⁡x))−cos⁡(2ln⁡x))+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⁡2θ=1−cos⁡2θ2{\sin ^{\mathbf{2}}}\theta = \dfrac{{1 - \cos 2\theta }}{2}, students should take care of changing the angles, as we can’t use ∠θ\angle \theta , we have to use ln⁡x,\ln x,because that’s how the question is given. And students should also take care that here we have taken ln⁡x=t\ln x = t, but the question is given in terms of x, so we need to get the answers in terms of x only.