Question

π/4∫ cosx/√2-sin²x dx ,Evaluate it.0

Answer 1

evaluate $\int\limits^{\pi/4}_0{\frac{cosx}{\sqrt{2-sin^2x}}}\,dx$

This question can be solved easily with help of substitution method.

let sinx = t

differentiating both sides,

cosx dx = dt......(1)

upper limit : sinπ/4 = 1/√2

lower limit : sin0 = 1

then, $\int\limits^{\pi/4}_0{\frac{cosx}{\sqrt{2-sin^2x}}}\,dx$ converts into $\int\limits^{1/\sqrt{2}}_0{\frac{1}{\sqrt{2-t^2}}}\,dt$

we know,

$\int{\frac{1}{\sqrt{a^2-x^2}}}\,dx=sin^{-1}\left(\frac{x}{a}\right)+C$

so, $\int\limits^{1/\sqrt{2}}_0{\frac{1}{\sqrt{2-t^2}}}\,dt$

= $\left[sin^{-1}\left(\frac{x}{\sqrt{2}}\right)\right]^{1/\sqrt{2}}_0$

= π/6