Question

Prove that:∫(√tanx+√cotx)dx=√2.π/2 for x→0,π/4

Answer 1

HELLO DEAR,

$\bold{I = \int\limits^{\frac{\pi}{4}}_0 {(\sqrt{tanx} + \sqrt{cotx})} \, dx }$

$\bold{I = \int\limits^{\frac{\pi}{4}}_0 {(\frac{\sqrt{sinx}}{\sqrt{cosx}} + \frac{\sqrt{cosx}}{\sqrt{sinx}})}\,dx}$

$\bold{I = \int\limits^{\frac{\pi}{4}}_0 {(\frac{sinx + cosx}{\sqrt{sinx*cosx}})}\,dx}$

$\bold{I = \sqrt{2}\int\limits^{\frac{\pi}{4}}_0 {\frac{sinx + cosx}{\sqrt{2sinxcosx}}}\,dx}$

$\bold{I = \sqrt{2}\int\limits^{\frac{\pi}{4}}_0 {\frac{sinx + cosx}{\sqrt{sin2x}}}\,dx}$

$\bold{I = \sqrt{2}\int\limits^{\frac{\pi}{4}}_0 {\frac{sinx + cosx}{\sqrt{1 - (1 - sin2x)}}}\,dx}$

$\bold{I = \sqrt{2}\int\limits^{\frac{\pi}{4}}_0 {\frac{sinx + cosx}{\sqrt{1 - (sinx - cosx)^2}}}\,dx}$

let t = (sinx - cosx)

${\Rightarrow }$ dt = (cosx + sinx).dxAlso, [x = 0 ${\Rightarrow }$ t = -1] and [x = π/4 ${\Rightarrow }$t = 0 ]

$\bold{\therefore, I = \sqrt{2}\int\limits^{0}_{-1} {\frac{dt}{\sqrt{1 - t^2}}}}$

$\bold{\Rightarrow I = \sqrt{2}[sin^{-1}t]^{0}_{-1}}$

$\bold{\Rightarrow I = \sqrt{2}[sin^{-1}(0) - sin^{-1}(-1)]}$

$\bold{\Rightarrow I = \sqrt{2}[0 - (-\pi/2)]}$

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

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

I HOPE ITS HELP YOU DEAR,

THANKS