If \( \int\displaystyle\frac{1+\cos 8\theta}{\tan 2\theta-\cot 2\theta}\,d\theta=\lambda\cos 8\theta+c \), then \( \lambda \) is equal to (where \( c \) is constant of integration)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{1}{16} \)
The numerator and the denominator both simplify in terms of \( 4\theta \):
\( 1+\cos 8\theta=2\cos^2 4\theta \)
\( \displaystyle \tan 2\theta-\cot 2\theta=\frac{\sin^2 2\theta-\cos^2 2\theta}{\sin 2\theta\cos 2\theta} \)
\( \displaystyle =\frac{-\cos 4\theta}{\frac{1}{2}\sin 4\theta} \)
So the integrand is
\( \displaystyle \frac{2\cos^2 4\theta\cdot\sin 4\theta}{-2\cos 4\theta} \)
\( =-\sin 4\theta\cos 4\theta \)
\( \displaystyle =-\frac{1}{2}\sin 8\theta \)
Integrating:
\( \displaystyle \int-\frac{1}{2}\sin 8\theta\,d\theta=\frac{\cos 8\theta}{16}+c \)
So \( \lambda=\frac{1}{16} \).