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Find : ∫ x √(1 + 2x) dx
Evaluate : ∫₀^(π²/4) (sin √x)/√x dx
Evaluate : ∫₀^π (sin x + cos x)/(9 + 16 sin 2x) dx
Evaluate : ∫₀^(π/2) sin 2x tan⁻¹(sin x) dx
∫₀^(π/2) (sin x – cos x)/(1 + sin x cos x) dx is equal to :
The value of ∫₀³ dx/√(9–x²) is:
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