Question

π/2∫ √1-sin 2x dx ,Evaluate it.π/4

Answer 1

Evaluate $\int\limits^{\pi/2}_{\pi/4}{\sqrt{1-sin2x}}\,dx$

first resolve √(1 - sin2x) into reasonable format.

we know, sin2θ = 2sinθcosθ

so, sin2x = 2sinx.cosx

also we can write 1 = sin²x + cos²x

because we know, sin²θ + cos²θ = 1

hence, √(1 - sin2x) = √{sin²x + cos²x - 2sinx.cosx}

= √{(sinx - cosx)²}

= |sinx - cosx|

we know, in π/4 to π/2

sinx > cosx

so, |sinx - cosx | = (sinx - cosx) in interval (π/4, π/2)

so, $\int\limits^{\pi/2}_{\pi/4}{(sinx-cosx)}\,dx$

= $\left[-cosx-sinx\right]^{\pi/2}_{\pi/4}$

= -[{cosπ/2 + sinπ/2}-{cosπ/4 + sinπ/4}]

= -[1 - √2]

= √2 - 1