\( \int\displaystyle\frac{1+\cos\theta}{1-\cos\theta}d\theta= \)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( -2\cot\left(\displaystyle\frac{\theta}{2}\right)-\theta+c \), where \( c \) is an arbitrary constant
By the half-angle identities \( 1+\cos\theta=2\cos^2\frac{\theta}{2} \) and \( 1-\cos\theta=2\sin^2\frac{\theta}{2} \),
\( \displaystyle \frac{1+\cos\theta}{1-\cos\theta}=\frac{2\cos^2\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}} \)
\( \displaystyle =\cot^2\frac{\theta}{2} \)
\( \displaystyle =\mathrm{cosec}^2\frac{\theta}{2}-1 \)
Since \( \int\mathrm{cosec}^2(kx)\,dx=-\frac{1}{k}\cot(kx) \) with \( k=\frac{1}{2} \),
\( \displaystyle \int\left(\mathrm{cosec}^2\frac{\theta}{2}-1\right)d\theta=-2\cot\left(\frac{\theta}{2}\right)-\theta+c \)
Shortcut: Differentiate Option (4):
\( \displaystyle -2\cdot\left(-\mathrm{cosec}^2\frac{\theta}{2}\right)\cdot\frac{1}{2}-1 \)
\( \displaystyle =\mathrm{cosec}^2\frac{\theta}{2}-1=\cot^2\frac{\theta}{2} \), the integrand.