Let \(I = \int_{0}^{\frac{\pi}{2}}\displaystyle\frac{\sin x}{\sin x + \cos x}\,dx\). Which of the following statements are correct?
(A) \(I = \int_{0}^{\frac{\pi}{2}}\displaystyle\frac{\cos x}{\sin x + \cos x}\,dx\)
(B) \(I = \displaystyle\frac{\pi}{4}\)
(C) \(\int_{0}^{\frac{\pi}{2}}\displaystyle\frac{\sin x - \cos x}{1 + \sin x\cos x}\,dx = \displaystyle\frac{\pi}{2}\)
(D) \(\int_{0}^{\frac{\pi}{2}}\displaystyle\frac{1}{1 + \tan x}\,dx = \displaystyle\frac{\pi}{2}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (B) only
By the property \(\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx\) with \(a = \frac{\pi}{2}\), replacing x by \(\frac{\pi}{2} - x\) swaps sin x and cos x:
\( \displaystyle I = \int_{0}^{\frac{\pi}{2}}\frac{\cos x}{\cos x + \sin x}\,dx \)
Adding the two forms of I:
\( \displaystyle 2I = \int_{0}^{\frac{\pi}{2}}\frac{\sin x + \cos x}{\sin x + \cos x}\,dx = \frac{\pi}{2} \)
\( \displaystyle I = \frac{\pi}{4} \)
The same swap turns \(\frac{\sin x - \cos x}{1 + \sin x\cos x}\) into its negative, so that integral J satisfies
\( J = -J \)
\( J = 0 \)
Also, dividing numerator and denominator by \(\cos x\),
\( \displaystyle \frac{1}{1 + \tan x} = \frac{\cos x}{\cos x + \sin x} \)
So \(I = \frac{\pi}{4}\), the integral in (C) is 0, and the integral in (D) equals I.
Checking the statements
(A) \(I = \int_{0}^{\frac{\pi}{2}}\frac{\cos x}{\sin x + \cos x}\,dx\) — Correct. Replacing x by \(\frac{\pi}{2} - x\) gives this form.
(B) \(I = \frac{\pi}{4}\) — Correct. 2I is the integral of 1 over \([0, \frac{\pi}{2}]\).
(C) \(\int_{0}^{\frac{\pi}{2}}\frac{\sin x - \cos x}{1 + \sin x\cos x}\,dx = \frac{\pi}{2}\) — Incorrect. The swap changes the sign of the integrand, so the value is 0.
(D) \(\int_{0}^{\frac{\pi}{2}}\frac{1}{1 + \tan x}\,dx = \frac{\pi}{2}\) — Incorrect. \(\frac{1}{1 + \tan x} = \frac{\cos x}{\cos x + \sin x}\), so the value is \(\frac{\pi}{4}\).