JEE Maths›Definite & Indefinite Integrals›Visual Solution
Visual SolutionPYQ 2023 · Apr Shift 1Standard
Question
The value of ∫0π/2sin⁡2x1+sin⁡xcos⁡x dx\int_0^{\pi/2} \frac{\sin^2 x}{1 + \sin x \cos x} \, dx is:
(A)π33\frac{\pi}{3\sqrt{3}}
(B)π23\frac{\pi}{2\sqrt{3}}
(C)π3\frac{\pi}{\sqrt{3}}
(D)2π33\frac{2\pi}{3\sqrt{3}}
Solution Path
King's property swaps sin⁡2x→cos⁡2x\sin^2 x \to \cos^2 x, add both forms →2I=∫dx1+sin⁡xcos⁡x\to 2I = \int \frac{dx}{1 + \sin x \cos x}, Weierstrass sub gives π33\frac{\pi}{3\sqrt{3}}
01Question Setup
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Evaluate I=∫0π/2sin⁡2x1+sin⁡xcos⁡x dxI = \int_0^{\pi/2} \frac{\sin^2 x}{1 + \sin x \cos x} \, dx. The mixed sin⁡xcos⁡x\sin x \cos x in the denominator suggests using King's property.
I=∫0π/2sin⁡2x1+sin⁡xcos⁡x dxI = \int_0^{\pi/2} \frac{\sin^2 x}{1 + \sin x \cos x} \, dx
02Apply King's PropertyKEY INSIGHT
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Replace xx with π/2−x\pi/2 - x: sin⁡x→cos⁡x\sin x \to \cos x, cos⁡x→sin⁡x\cos x \to \sin x. The denominator is unchanged, so II also equals ∫0π/2cos⁡2x1+sin⁡xcos⁡x dx\int_0^{\pi/2} \frac{\cos^2 x}{1 + \sin x \cos x} \, dx.
I=∫0π/2cos⁡2x1+sin⁡xcos⁡x dxI = \int_0^{\pi/2} \frac{\cos^2 x}{1 + \sin x \cos x} \, dx (by King's property)
03Add Both Forms
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Adding the two forms: 2I=∫0π/211+sin⁡xcos⁡xdx=∫0π/222+sin⁡2xdx2I = \int_0^{\pi/2} \frac{1}{1 + \sin x \cos x} dx = \int_0^{\pi/2} \frac{2}{2 + \sin 2x} dx. Substitute t=tan⁡xt = \tan x and complete the square: ∫0∞dtt2+t+1\int_0^{\infty} \frac{dt}{t^2 + t + 1}.
2I=∫0∞dt(t+1/2)2+3/4=2π332I = \int_0^{\infty} \frac{dt}{(t + 1/2)^2 + 3/4} = \frac{2\pi}{3\sqrt{3}}
04Final Answer
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From 2I=2π332I = \frac{2\pi}{3\sqrt{3}}, we get I=π33I = \frac{\pi}{3\sqrt{3}}.
I=π33I = \frac{\pi}{3\sqrt{3}}, Answer: (A)
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King's Property

EASY

For definite integrals, replace x with (a+b-x) to exploit symmetry

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

If f(a+b-x) = -f(x), the integral is zero. If f(a+b-x) = f(x), the integral doubles

Symmetry integralsTrigonometric integralsPolynomial integrals
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Substitution Method

EASY

Replace a composite expression with a single variable to simplify the integrand

∫f(g(x)) g′(x) dx=∫f(t) dt,t=g(x)\int f(g(x))\,g'(x)\,dx = \int f(t)\,dt, \quad t = g(x)

Transforms complex integrands into standard forms that can be integrated directly

Trigonometric integralsRational functionsPower rule integrals
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