Integrals: Tricky Questions Solved (3)

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Tags Integrals

Question

Evaluate ∫0π/2sin⁡xsin⁡x+cos⁡x dx\displaystyle \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx. This is a JEE Main classic that tests the king’s property of definite integrals.

Solution — Step by Step

For any continuous ff on [a,b][a, b]:

∫abf(x) dx=∫abf(a+b−x) dx\int_a^b f(x)\, dx = \int_a^b f(a + b - x)\, dx

For a=0a = 0, b=π/2b = \pi/2: replace xx with π/2−x\pi/2 - x. Useful because sin⁡(π/2−x)=cos⁡x\sin(\pi/2 - x) = \cos x and cos⁡(π/2−x)=sin⁡x\cos(\pi/2 - x) = \sin x.

Let

I=∫0π/2sin⁡xsin⁡x+cos⁡x dxI = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\, dx

Replacing x→π/2−xx \to \pi/2 - x:

I=∫0π/2cos⁡xcos⁡x+sin⁡x dxI = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}\, dx

2I=∫0π/2sin⁡x+cos⁡xsin⁡x+cos⁡x dx=∫0π/21 dx=π22I = \int_0^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}\, dx = \int_0^{\pi/2} 1\, dx = \frac{\pi}{2}

I=π4I = \frac{\pi}{4}

Why This Works

The king’s property exploits symmetry: when an integrand has a “partner” form related by x→a+b−xx \to a+b-x, adding the two expressions often collapses to something trivial. Here, sin⁡x\sqrt{\sin x} and cos⁡x\sqrt{\cos x} are partners, and adding II to its transformed self gives a constant integrand.

This is one of the four “magic tricks” of definite integrals (along with the property that integral over symmetric interval of an odd function is zero, the integral of an even function is twice the half-integral, and the periodicity property). JEE loves it.

Alternative Method

You could try direct substitution u=sin⁡xu = \sin x or u=tan⁡xu = \tan x, but they lead to gnarly elliptic-style integrals with no closed form. The king’s property is the only practical method here.

Whenever you see sin⁡x/(sin⁡x+cos⁡x)\sqrt{\sin x}/(\sqrt{\sin x} + \sqrt{\cos x}) or similar “ratio with sin⁡/cos⁡\sin/\cos partner” structure on [0,π/2][0, \pi/2], the answer is almost always π/4\pi/4. Recognise this template and you can guess the answer in a few seconds, then verify with the king’s property.

Common Mistake

Two errors plague this template:

  1. Wrong substitution. Some students try x→π−xx \to \pi - x (Western half-angle) or x→−xx \to -x (symmetric interval). Neither works on [0,π/2][0, \pi/2] — the right move is x→π/2−xx \to \pi/2 - x because that’s the limits’ midpoint reflection.

  2. Forgetting that the new integral has the same limits. When you substitute u=π/2−xu = \pi/2 - x, du=−dxdu = -dx, but the limits also flip (x=0→u=π/2x = 0 \to u = \pi/2, x=π/2→u=0x = \pi/2 \to u = 0). The two negatives cancel, and the integral keeps the same limits. Some students get confused and pick up an extra negative.

Final answer: I=π4I = \dfrac{\pi}{4}.

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