Integrals: Conceptual Doubts Cleared (12)

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Tags Integrals

Question

Evaluate ∫0π/2sin⁡2xsin⁡x+cos⁡x dx\int_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} \, dx using the property ∫0af(x) dx=∫0af(a−x) dx\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx.

Solution — Step by Step

Let I=∫0π/2sin⁡2xsin⁡x+cos⁡x dxI = \int_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} \, dx. Using x→π/2−xx \to \pi/2 - x:

I=∫0π/2cos⁡2xcos⁡x+sin⁡x dxI = \int_0^{\pi/2} \frac{\cos^2 x}{\cos x + \sin x} \, dx 2I=∫0π/2sin⁡2x+cos⁡2xsin⁡x+cos⁡x dx=∫0π/21sin⁡x+cos⁡x dx2I = \int_0^{\pi/2} \frac{\sin^2 x + \cos^2 x}{\sin x + \cos x} \, dx = \int_0^{\pi/2} \frac{1}{\sin x + \cos x} \, dx

Use sin⁡x+cos⁡x=2sin⁡(x+π/4)\sin x + \cos x = \sqrt{2} \sin(x + \pi/4):

2I=12∫0π/2csc⁡(x+π/4) dx2I = \frac{1}{\sqrt{2}} \int_0^{\pi/2} \csc(x + \pi/4) \, dx

∫csc⁡u du=ln⁡∣tan⁡(u/2)∣+C\int \csc u \, du = \ln |\tan(u/2)| + C.

Evaluate from x=0x = 0 to π/2\pi/2, i.e., u=π/4u = \pi/4 to 3π/43\pi/4:

2I=12[ln⁡∣tan⁡(3π/8)∣−ln⁡∣tan⁡(π/8)∣]2I = \frac{1}{\sqrt{2}} [\ln |\tan(3\pi/8)| - \ln |\tan(\pi/8)|].

tan⁡(π/8)=2−1\tan(\pi/8) = \sqrt{2} - 1, tan⁡(3π/8)=2+1\tan(3\pi/8) = \sqrt{2} + 1. So:

2I=12ln⁡2+12−1=12ln⁡(2+1)2=22ln⁡(2+1)=2ln⁡(2+1)2I = \frac{1}{\sqrt{2}} \ln \frac{\sqrt{2}+1}{\sqrt{2}-1} = \frac{1}{\sqrt{2}} \ln(\sqrt{2}+1)^2 = \frac{2}{\sqrt{2}} \ln(\sqrt{2}+1) = \sqrt{2} \ln(\sqrt{2}+1)

So I=12ln⁡(2+1)I = \frac{1}{\sqrt{2}} \ln(\sqrt{2}+1).

Final answer: I=12ln⁡(1+2)I = \frac{1}{\sqrt{2}} \ln(1 + \sqrt{2}).

Why This Works

The “king” property ∫0af(x)dx=∫0af(a−x)dx\int_0^a f(x) dx = \int_0^a f(a-x) dx is a major weapon for symmetric integrals. By replacing xx with π/2−x\pi/2 - x, sin⁡x\sin x swaps with cos⁡x\cos x. Adding the two versions of II collapses sin⁡2+cos⁡2\sin^2 + \cos^2 to 1, killing the squared term.

This is the cleanest path through such integrals — far better than substitution or partial fractions, which would take 3x longer.

Alternative Method

Substitute t=tan⁡(x/2)t = \tan(x/2) (Weierstrass). sin⁡x=2t/(1+t2)\sin x = 2t/(1+t^2), cos⁡x=(1−t2)/(1+t2)\cos x = (1-t^2)/(1+t^2). Tedious but works. The king method is much faster.

Whenever you see ∫0π/2f(sin⁡x,cos⁡x)dx\int_0^{\pi/2} f(\sin x, \cos x) dx where the integrand has visible sin⁡\sin/cos⁡\cos symmetry, try x→π/2−xx \to \pi/2 - x first. The identity sin⁡2+cos⁡2=1\sin^2 + \cos^2 = 1 does most of the heavy lifting.

Common Mistake

Forgetting that ∫0π/2sin⁡xsin⁡x+cos⁡xdx=π/4\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx = \pi/4 (a JEE classic). Students sometimes confuse this with our integral and short-circuit to a wrong answer.

Sign error in the cosec integral. Antiderivative of csc⁡u\csc u is −ln⁡∣csc⁡u+cot⁡u∣=ln⁡∣tan⁡(u/2)∣-\ln|\csc u + \cot u| = \ln|\tan(u/2)| — both forms are valid, just be consistent.

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