Question

Show that ∫(0->π/2) √(tanx)+√(cotx)=√2π

Answer 1

HELLO DEAR,

$\bold{I = \int\limits^{\frac{\pi}{2}}_0 {(\sqrt{tanx} + \sqrt{cotx})} \, dx }$

$\bold{I = \int\limits^{\frac{\pi}{2}}_0 {(\frac{\sqrt{sinx}}{\sqrt{cosx}} + \frac{\sqrt{cosx}}{\sqrt{sinx}})}\,dx}$

$\bold{I = \int\limits^{\frac{\pi}{2}}_0 {(\frac{sinx + cosx}{\sqrt{sinx*cosx}})}\,dx}$

$\bold{I = \sqrt{2}\int\limits^{\frac{\pi}{2}}_0 {\frac{sinx + cosx}{\sqrt{2sinxcosx}}}\,dx}$

$\bold{I = \sqrt{2}\int\limits^{\frac{\pi}{2}}_0 {\frac{sinx + cosx}{\sqrt{sin2x}}}\,dx}$$\bold{I = \sqrt{2}\int\limits^{\frac{\pi}{2}}_0 {\frac{sinx + cosx}{\sqrt{1 - (1 - sin2x)}}}\,dx}$

$\bold{I = \sqrt{2}\int\limits^{\frac{\pi}{2}}_0 {\frac{sinx + cosx}{\sqrt{1 - (sinx - cosx)^2}}}\,dx}$

let t = (sinx - cosx)
${\Rightarrow }$ dt = (cosx + sinx).dx
Also, [x = 0 ${\Rightarrow }$ t = -1] and [x = π/2 ${\Rightarrow }$t = 1 ]

$\bold{\therefore, I = \sqrt{2}\int\limits^{1}_{-1} {\frac{dt}{\sqrt{1 - t^2}}}}$

$\bold{\Rightarrow I = \sqrt{2}[sin^{-1}t]^{1}_{-1}}$

$\bold{\Rightarrow I = \sqrt{2}[sin^{-1}(1) - sin^{-1}(-1)]}$

$\bold{\Rightarrow I = \sqrt{2}[\pi/2 - (-\pi/2)]}$

$\bold{\Rightarrow I = \sqrt{2}*(\pi/2 + \frac{\pi}{2})}$

$\bold{\Rightarrow I = \sqrt{2}\pi}$

I HOPE ITS HELP YOU DEAR,
THANKS