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Indefinite Integrals

Section: Mathematics

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Question : 4 of 90

Marks: +1, -0

Let

f(t)=\int \left( \frac{1- \sin (\log_e t)}{1-\cos(\log_e t)} \right) dt, t>1

f\left(e^{\frac{\pi}{2}}\right) = -e^{\frac{\pi}{2}}

f\left(e^{\frac{\pi}{4}}\right) = \alpha e^{\frac{\pi}{4}}

\alpha

[24 jan 2026 Shift 1]

$1+\sqrt{2}$
$-1-2\sqrt{2}$
$-1-\sqrt{2}$
$-1+\sqrt{2}$

Solution:

Concept:

Integration of the form

\int e^x [g(x) + g'(x)]\,dx = e^x g(x) + C

.

Explanation:

Let

\log_e t = x

.

Then

t = e^x

and

dt = e^x\,dx

.

The integral becomes

f(t) = \int \frac{1-\sin x}{1-\cos x} e^x\,dx

.

Use half‑angle identities:

1-\cos x = 2\sin^2\frac{x}{2}

,

\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}

.

So

\frac{1-\sin x}{1-\cos x} = \frac{1}{2}\csc^2\frac{x}{2} - \cot\frac{x}{2}

.

Thus

f(t) = \int e^x \left(\frac{1}{2}\csc^2\frac{x}{2} - \cot\frac{x}{2}\right)dx

.

Let

g(x) = -\cot\frac{x}{2}

. Then

g'(x) = \frac{1}{2}\csc^2\frac{x}{2}

.

Hence the integrand is

e^x[g(x)+g'(x)]

and the integral is

e^x g(x) + C = -e^x\cot\frac{x}{2}+C

.

Substitute back

x=\log_e t

:

f(t) = -t\cot\left(\frac{\log_e t}{2}\right) + C

.

Use

f(e^{\pi/2}) = -e^{\pi/2}

:

-e^{\pi/2} = -e^{\pi/2}\cot(\pi/4)+C = -e^{\pi/2}+C

, so

C=0

.

Thus

f(t) = -t\cot\left(\frac{\log_e t}{2}\right)

.

Now

\alpha e^{\pi/4} = f(e^{\pi/4}) = -e^{\pi/4}\cot(\pi/8)

→

\alpha = -\cot(\pi/8)

.

Use

\cot\theta = \frac{1+\cos2\theta}{\sin2\theta}

with

\theta = \pi/8

→

\cot(\pi/8) = \frac{1+\cos(\pi/4)}{\sin(\pi/4)} = \frac{1+1/\sqrt{2}}{1/\sqrt{2}} = \sqrt{2}+1

.

So

\alpha = -(\sqrt{2}+1) = -1-\sqrt{2}

.

Answer:

Option C:

-1-\sqrt{2}

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