Question

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. $1+\sqrt{2}$
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Hint:

Always convert logarithmic arguments using substitution to simplify exponential integrals.

Answer 1

The Correct Option is D

Step 1: Substitution.
Let \[ x=\log_e t \Rightarrow dt=e^x dx \] \[ f(t)=\int \frac{1-\sin x}{1-\cos x}e^x dx \] Using \[ \frac{1-\sin x}{1-\cos x}=\frac{(1-\sin x)(1+\cos x)}{(1-\cos x)(1+\cos x)} =\frac{(1-\sin x)(1+\cos x)}{\sin^2 x} \] Simplifying, \[ =1+\csc x-\cot x \] Step 2: Integrating.
\[ f(t)=\int e^x(1+\csc x-\cot x)\,dx \] \[ = e^x(1+\csc x)+C \] Step 3: Using $f(e^{\pi/2})=-e^{\pi/2$.}
\[ -e^{\pi/2}=e^{\pi/2}(1+\csc \frac{\pi}{2})+C \] \[ -e^{\pi/2}=2e^{\pi/2}+C \Rightarrow C=-3e^{\pi/2} \] Step 4: Evaluating $f(e^{\pi/4})$.
\[ f(e^{\pi/4})=e^{\pi/4}\left(1+\sqrt{2}\right) \] \[ \alpha=1+\sqrt{2} \]