Correct Answer is : x⋅tan⁡(log⁡x2)+cx · ≤ft( x/2)+cx⋅tan(2logx)+c, where c is a constant of integration.

x² ≤ft( x/2)+c, where c is a constant of integration.

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

x³ ≤ft( x/2)+c, where c is a constant of integration.

x · ≤ft( x/2)+c, where c is a constant of integration.

Correct Answer is : x⋅tan⁡(log⁡x2)+cx · ≤ft( x/2)+cx⋅tan(2logx)+c, where c is a constant of integration.

Test Index

MHTCET 10 May 2024 Shift 1 Paper

Question : 133 of 150

Marks: +1, -0

\int \frac{1+\sin (\log x)}{1+\cos (\log x)} d x=

$x^{2} \tan \left(\frac{\log x}{2}\right)+c$ , where $c$ is a constant of integration.
$x \tan \left(\log \left(\frac{x}{2}\right)\right)+c$ , where c is a constant of integration.
$x^{3} \log \left(\frac{\tan x}{2}\right)+c$ , where c is a constant of integration.
$x \cdot \tan \left(\frac{\log x}{2}\right)+c$ , where c is a constant of integration.

Solution:

Concept:

Substitute

\log x = t

and use half-angle identities to reduce the integrand to the form

e^t(f(t)+f'(t))

Explanation:

Let

t = \log x

, hence

x = e^t

and

dx = e^t dt

The integral becomes

I = \int \frac{1+\sin t}{1+\cos t} e^t dt

Using

1+\cos t = 2\cos^2(t/2)

and

\sin t = 2\sin(t/2)\cos(t/2)

, we get:

1+\sin t = \sin^2(t/2)+\cos^2(t/2)+2\sin(t/2)\cos(t/2) = \left(\sin(t/2)+\cos(t/2)\right)^2

Therefore,

\frac{1+\sin t}{1+\cos t} = \frac{1}{2}\left(\sec^2(t/2)+2\tan(t/2)\right)

I = \frac{1}{2}\int e^t\left[\sec^2(t/2)+2\tan(t/2)\right]dt

Observe that

\frac{d}{dt}\left[\tan(t/2)\right] = \frac{1}{2}\sec^2(t/2)

Thus the integrand becomes

e^t\left[f(t)+f'(t)\right]

with

f(t)=\tan(t/2)

Using

\int e^t\left[f(t)+f'(t)\right]dt = e^t f(t)+c

, we have:

I = e^t \tan(t/2)+c

Substitute back

t = \log x

and

e^t = x

I = x\tan\left(\frac{\log x}{2}\right)+c

Answer:

x\tan\left(\frac{\log x}{2}\right)+c

, which matches option D.

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