MHT CET · Maths · Definite Integration
\(\int_{0}^{\pi / 2} \frac{d x}{1+\tan x}\) is equal to
- A \(\pi\)
- B \(\pi / 2\)
- C \(\pi / 3\)
- D \(\pi / 4\)
Answer & Solution
Correct Answer
(D) \(\pi / 4\)
Step-by-step Solution
Detailed explanation
Given, \(I=\int_{0}^{\pi / 2} \frac{d x}{1+\tan x}\)
\(I =\int_{0}^{\pi / 2} \frac{\cos x}{\sin x+\cos x} d x \ldots \)
\( I =\int_{0}^{\pi / 2} \frac{\cos (\pi / 2-x)}{\sin (\pi / 2-x)+\cos (\pi / 2-x)} d x \)
\( =\int_{0}^{\pi / 2} \frac{\sin x}{\cos x+\sin x} d x \ldots\)
On adding Eqs. (i) and (ii), we get
\(2 I =\int_{0}^{\pi / 2}\left(\frac{\sin x+\cos x}{\sin x+\cos x}\right) d x \)
\( =\int_{0}^{\pi / 2} d x=\pi / 2 \)
\( \Rightarrow \quad I =\pi / 4\)
\(I =\int_{0}^{\pi / 2} \frac{\cos x}{\sin x+\cos x} d x \ldots \)
\( I =\int_{0}^{\pi / 2} \frac{\cos (\pi / 2-x)}{\sin (\pi / 2-x)+\cos (\pi / 2-x)} d x \)
\( =\int_{0}^{\pi / 2} \frac{\sin x}{\cos x+\sin x} d x \ldots\)
On adding Eqs. (i) and (ii), we get
\(2 I =\int_{0}^{\pi / 2}\left(\frac{\sin x+\cos x}{\sin x+\cos x}\right) d x \)
\( =\int_{0}^{\pi / 2} d x=\pi / 2 \)
\( \Rightarrow \quad I =\pi / 4\)
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