Correct Answer is : x2(cos⁡(log⁡x)+sin⁡(log⁡x))+cx/2( ( x)+ ( x))+c2x(cos(logx)+sin(logx))+c, (where c is a constant of integration)

x/2( ( x)- ( x))+c_{l} (where c is a constant of integration)

x( ( x)- ( x))+c_{c} (where c is a constant of integration)

x/2( ( x)+ ( x))+c, (where c is a constant of integration)

x( ( x)+ ( x))+c₁ (where c is a constant of integration)

Correct Answer is : x2(cos⁡(log⁡x)+sin⁡(log⁡x))+cx/2( ( x)+ ( x))+c2x(cos(logx)+sin(logx))+c, (where c is a constant of integration)

Test Index

MHTCET 11 May 2024 Shift 2 Paper

Question : 128 of 150

Marks: +1, -0

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

$\frac{x}{2}(\sin (\log x)-\cos (\log x))+c_{l}$ (where c is a constant of integration)
$x(\cos (\log x)-\sin (\log x))+c_{c}$ (where c is a constant of integration)
$\frac{x}{2}(\cos (\log x)+\sin (\log x))+c$ , (where c is a constant of integration)
$x(\cos (\log x)+\sin (\log x))+c_{1}$ (where c is a constant of integration)

Solution:

Concept:

Use substitution followed by the integration by parts method to evaluate the integral.

Explanation:

Let

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

Put

\log x = t

. Then

x = e^t

and

dx = e^t\, dt

I = \int e^t \cos t\, dt

Apply integration by parts:

I = e^t \cos t - \int (-\sin t) e^t\, dt = e^t\cos t + \int e^t \sin t\, dt

Again integrate by parts:

\int e^t \sin t\, dt = e^t \sin t - \int e^t \cos t\, dt = e^t \sin t - I

Thus,

I = e^t \cos t + e^t \sin t - I

2I = e^t(\cos t + \sin t)

Substitute

e^t = x

and

t = \log x

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

Answer:

\frac{x}{2}(\cos(\log x) + \sin(\log x)) + c

, which matches Option C.

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