\(\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\displaystyle\frac{dx}{1+\cos x}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
By the half-angle identity \(1+\cos x = 2\cos^{2}\frac{x}{2}\),
\( \displaystyle \frac{1}{1+\cos x} = \frac{1}{2}\sec^{2}\frac{x}{2} \)
An antiderivative of \(\frac{1}{2}\sec^{2}\frac{x}{2}\) is \(\tan\frac{x}{2}\). So
\( \displaystyle \int_{\pi/4}^{3\pi/4}\frac{dx}{1+\cos x} = \tan\frac{3\pi}{8} - \tan\frac{\pi}{8} \)
Using \(\tan\frac{\pi}{8} = \sqrt{2}-1\) and \(\tan\frac{3\pi}{8} = \sqrt{2}+1\):
\( (\sqrt{2}+1) - (\sqrt{2}-1) = 2\)