Integrals: Speed-Solving Techniques (8)

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Tags Integrals

Question

Evaluate ∫0π/2sin⁡xsin⁡x+cos⁡x dx\displaystyle \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\, dx using a king’s-property shortcut.

Solution — Step by Step

For definite integrals on [0,a][0, a]:

∫0af(x) dx=∫0af(a−x) dx\int_0^a f(x)\, dx = \int_0^a f(a - x)\, dx

With a=π/2a = \pi/2, replace x→π/2−xx \to \pi/2 - x. Note 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.

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

Let original I=∫0π/2sin⁡xsin⁡x+cos⁡x dxI = \displaystyle \int_0^{\pi/2} \dfrac{\sin x}{\sin x + \cos x}\, dx and the new form I′=∫0π/2cos⁡xsin⁡x+cos⁡x dxI' = \displaystyle \int_0^{\pi/2} \dfrac{\cos x}{\sin x + \cos x}\, dx. By King’s property, I=I′I = I'. So:

2I=∫0π/2sin⁡x+cos⁡xsin⁡x+cos⁡x dx=∫0π/21 dx=π22I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x}\, dx = \int_0^{\pi/2} 1\, dx = \frac{\pi}{2} I=π4I = \frac{\pi}{4}

Final Answer: I=π4I = \dfrac{\pi}{4}.

Why This Works

King’s property exploits the symmetry f(x)+f(a−x)f(x) + f(a - x) to collapse a complicated rational trig integral into a constant. The full method takes three lines once you spot the substitution; brute-force integration of sin⁡x/(sin⁡x+cos⁡x)\sin x / (\sin x + \cos x) would take half a page using Weierstrass or partial-fraction tricks.

The key clue: whenever the denominator is symmetric in sin⁡\sin and cos⁡\cos but the numerator is asymmetric, King’s property usually closes the problem.

Alternative Method

Multiply numerator and denominator by 1/cos⁡x1/\cos x to get tan⁡xtan⁡x+1\dfrac{\tan x}{\tan x + 1}. Substitute t=tan⁡xt = \tan x — but the resulting integral ∫t dt(t+1)(1+t2)\int \dfrac{t \, dt}{(t+1)(1+t^2)} requires partial fractions. King’s property is faster.

Forgetting to write down the new integral I′I' as a separate object before adding leads to confusion about which integral is which. Always label both forms; then 2I=I+I′2I = I + I' falls out cleanly.

Memorize: any integral of the form ∫0π/2f(sin⁡x)f(sin⁡x)+f(cos⁡x) dx=π4\displaystyle \int_0^{\pi/2} \dfrac{f(\sin x)}{f(\sin x) + f(\cos x)}\, dx = \dfrac{\pi}{4}. This pattern alone solves at least 3 JEE Main MCQs every year.

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