Question

√2 ∫ √2-x² dx ,Evaluate it.0

Answer 1

we have to evaluate , $\int\limits^{\sqrt{2}}_0{\sqrt{2-x^2}}\,dx$

putting, x = √2sinθ

differentiating both sides,

dx = √2cosθ. dθ

then, √(2 - x²) = √{2 - (√2sinθ)²}

= √2√(1 - sin²θ)

= √2√(cos²θ) = √2cosθ

now, upper limit : √2 = √2sinθ ⇒x = π/2

and lower limit : 0 = √2sinθ ⇒x = 0

so, $\int\limits^{\sqrt{2}}_0{\sqrt{2-x^2}}\,dx$ converts into

$\int\limits^{\pi/2}_0{\sqrt{2}cos\theta}\,\sqrt{2}cos\theta d\theta$

= $2\int\limits^{\pi/2}_0{cos^2\theta}\,d\theta$

we know, cos²x = (1 + cos2x)/2

so, cos²θ = (1 + cos2θ)/2

$\textbf{then,}\int\limits^{\pi/2}_0{(1+cos2\theta)}\,d\theta=\left[\theta+\frac{sin2\theta}{2}\right]^{\pi/2}_0$

= π/2