Correct Answer is : x2[sin⁡(log⁡x)−cos⁡(log⁡x)]+cx/2[ ( x)- ( x)]+c2x[sin(logx)−cos(logx)]+c

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MHTCET 18 Apr 2026 Shift 1 Paper

Question : 129 of 150

Marks: +1, -0

\int \sin (\log x) d x=

Solution:

Concept:

Use integration by parts twice and then solve for the original integral.

Explanation:

Let

I = \int \sin(\log x)\,dx

Here

\log x

is the natural logarithm.

Integrate by parts taking

u = \sin(\log x)

and

dv = dx

Then

du = \frac{\cos(\log x)}{x}\,dx

and

v = x

So,

I = x\sin(\log x) - \int \cos(\log x)\,dx

Let

J = \int \cos(\log x)\,dx

Integrate

J

by parts with

u = \cos(\log x)

and

dv = dx

Then

du = -\frac{\sin(\log x)}{x}\,dx

and

v = x

Thus,

J = x\cos(\log x) + \int \sin(\log x)\,dx = x\cos(\log x) + I

Substitute

J

in the expression for

I

I = x\sin(\log x) - [x\cos(\log x) + I]

2I = x[\sin(\log x) - \cos(\log x)]

I = \frac{x}{2}[\sin(\log x) - \cos(\log x)] + C

Answer:

Option A:

\frac{x}{2}[\sin(\log x) - \cos(\log x)] + C

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