Let f(t)= ( 1- ( e t)1- ( e t))dt, t1. If f(e^ /2)=-e^ /2 and f(e^ /4)= e^ /4, then equals

Given: f(t)= ( 1- ( e t)1- ( e t))dt, t1 with f(e^ /2)=-e^ /2 and f(e^ /4)= e^ /4 Find: Use the substitution x= e t dt=e^x dx Then f(t)= 1- x1- xe^x dx Now simplify the trigonometric factor: 1- x1- x= (1- x)(1+ x)(1- x)(1+ x)= (1- x)(1+ x) ^2 x=1+ x- x Therefore, f(t)= e^x(1+ x- x) dx Hence, f(t)=e^x(1+ x)+C Using f(e^ /2)=-e^ /2, put x= 2: -e^ /2=e^ /2 (1+ 2)+C Since 2=1, -e^ /2=2e^ /2+C So, C=-3e^ /2 Now evaluate at t=e^ /4: f(e^ /4)=e^ /4 (1+ 2) Thus, =1+ 2 Therefore, the correct option is D. The solution concludes after substitution and simplification that f(t)=e^x(1+ x)+C with x= e t. Substituting x= 4 gives 4= 2, so f(e^ /4)=e^ /4(1+ 2) Comparing with f(e^ /4)= e^ /4, we get =1+ 2 So the correct option is D.

Let f(t)= ( 1- ( e t)1- ( e t))dt, t1. If f(e^ /2)=-e^ /2 and f(e… — Solution | JEE Mathematics

MCQMediumJEE 2026Indefinite Integrals

Let $f(t)=\int \left(\frac{1-\sin(\log_e t)}{1-\cos(\log_e t)}\right)dt,\; t>1.$ If $f(e^{\pi/2})=-e^{\pi/2}$ and $f(e^{\pi/4})=\alpha e^{\pi/4},$ then $\alpha$ equals

A
$-1+\sqrt{2}$
B
$-1-2\sqrt{2}$
C
$-1-\sqrt{2}$
D
$1+\sqrt{2}$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

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

with

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

and

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

Find: $\alpha$

Use the substitution

x=\log_e t \Rightarrow dt=e^x \, dx

Then

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

Now simplify the trigonometric factor:

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

Therefore,

f(t)=\int e^x(1+\csc x-\cot x) \, dx

Hence,

f(t)=e^x(1+\csc x)+C

Using $f(e^{\pi/2})=-e^{\pi/2}$ , put $x=\frac{\pi}{2}$ :

-e^{\pi/2}=e^{\pi/2}\left(1+\csc \frac{\pi}{2}\right)+C

Since $\csc \frac{\pi}{2}=1$ ,

-e^{\pi/2}=2e^{\pi/2}+C

So,

C=-3e^{\pi/2}

Now evaluate at $t=e^{\pi/4}$ :

f(e^{\pi/4})=e^{\pi/4}\left(1+\sqrt{2}\right)

Thus,

\alpha=1+\sqrt{2}

Therefore, the correct option is D.

Using the derived antiderivative

The solution concludes after substitution and simplification that

f(t)=e^x(1+\csc x)+C

with $x=\log_e t$ . Substituting $x=\frac{\pi}{4}$ gives $\csc \frac{\pi}{4}=\sqrt{2}$ , so

f(e^{\pi/4})=e^{\pi/4}(1+\sqrt{2})

Comparing with $f(e^{\pi/4})=\alpha e^{\pi/4}$ , we get

\alpha=1+\sqrt{2}

So the correct option is D.

Common mistakes

Using the substitution incorrectly. If $x=\log_e t$ , then $dt=e^x \, dx$ , not just $dx$ . Missing the factor $e^x$ changes the integral completely.
Simplifying $\frac{1-\sin x}{1-\cos x}$ incorrectly. The solution uses rationalization with $1+\cos x$ to obtain $1+\csc x-\cot x$ . Skipping this algebra leads to a wrong integrand.
Confusing the condition on $f(e^{\pi/2})$ with a definite integral limit. Here it is used to determine the constant of integration $C$ for the antiderivative.

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