Evaluate the integral from limit π/6 to π/3 sinx+cosx/root over sin 2x dx

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integrate sinx plus cos x upon under root sin2x from pi upon six to pi upon three wrt x

Find π/6 to π/3 int (sin x+ cos x)/√sin 2x dx | int(pi/6)^(pi/3) (sin x+ cos x)/(sqrt(sin 2x))dx

23) Evaluate π/3 to π/6 √sinx/√sinx+√cosx dx || Definite Integration || 2023 Regular TS

Calculus Help: Integral from (π/6) to (π/3) sinx/(sinx+cosx) dx - Integration by substitution - Trig

integrate (sin x + cos x)/(sqrt(sin 2x)) dx from pi/6 to pi/3

Integration:- Limit π/6 to π/3 sinx+cosx / sin2x^1/2

Evaluate:∫(π/6→π/3) (sinx+cosx)/√(sin2x) dx.

integrate sinx+cosx/√sin2x from pi/6 to pi/3 || int_(pi/6)^( pi/3)(sin x+cos x)/(sqrt(sin2x))dx