MHT CET · Maths · Definite Integration
\(\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{d x}{1+\cos x}\) is equal to
- A \(-2\)
- B \(-2-2 \sqrt{2}\)
- C \(2\)
- D \(-2 \sqrt{2}\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
\(\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{d x}{1+\cos x}=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{1}{2} \sec ^2 \frac{x}{2} d x=\left[\tan \frac{x}{2}\right]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \)
\( =\left(\tan \frac{3 \pi}{8}-\tan \frac{\pi}{8}\right)=\left(\cot \frac{\pi}{8}-\tan \frac{\pi}{8}\right)=\) \(\frac{\cos ^2 \frac{\pi}{8}-\sin ^2 \frac{\pi}{8}}{\sin \frac{\pi}{8} \cdot \cos \frac{\pi}{8}} \)
\( =\frac{2 \cos \frac{\pi}{4}}{\sin \frac{\pi}{4}}=2\)
\( =\left(\tan \frac{3 \pi}{8}-\tan \frac{\pi}{8}\right)=\left(\cot \frac{\pi}{8}-\tan \frac{\pi}{8}\right)=\) \(\frac{\cos ^2 \frac{\pi}{8}-\sin ^2 \frac{\pi}{8}}{\sin \frac{\pi}{8} \cdot \cos \frac{\pi}{8}} \)
\( =\frac{2 \cos \frac{\pi}{4}}{\sin \frac{\pi}{4}}=2\)
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