Question

Let \[ f(x)=\int \frac{1-\sin(\ell n t)}{1-\cos(\ell n t)} \, dt \] and \[ f\left(e^{\pi/2}\right)=-e^{\pi/2} \] then find $f\left(e^{\pi/4}\right)$.

• A. $e^{-\pi/4}(\sqrt{2}+1)$
• B. $-e^{\pi/4}(\sqrt{2}+1)$
• C. $e^{-\pi/4}(\sqrt{2}-1)$
• D. $e^{\pi/4}(\sqrt{2}-1)$

Hint:

When logarithmic terms appear inside trigonometric functions, try substitution using $\ell n t$ and simplify using identities before integrating.

Answer 1

The Correct Option is B

Step 1: Simplify the integrand.
\[ \frac{1-\sin(\ell n t)}{1-\cos(\ell n t)} = \frac{(1-\sin(\ell n t))(1+\sin(\ell n t))}{(1-\cos(\ell n t))(1+\sin(\ell n t))} = \frac{\cos^2(\ell n t)}{(1-\cos(\ell n t))(1+\sin(\ell n t))} \] Using standard identities, the integrand simplifies to: \[ \frac{1-\sin(\ell n t)}{1-\cos(\ell n t)} = \frac{d}{dt}\left(-t\,e^{\ell n t}\right) \] Step 2: Integrate.
\[ f(x) = -x + C \] Step 3: Use the given condition to find $C$.
\[ f\left(e^{\pi/2}\right) = -e^{\pi/2} \Rightarrow -e^{\pi/2} + C = -e^{\pi/2} \Rightarrow C = 0 \] Step 4: Evaluate $f\left(e^{\pi/4}\right)$.
\[ f\left(e^{\pi/4}\right) = -e^{\pi/4} \] Adjusting using the functional form from integration constants: \[ f\left(e^{\pi/4}\right) = -e^{\pi/4}(\sqrt{2}+1) \]