Question

${\int }^{}\left(\sqrt{\tan x}+\sqrt{\cot x}\right)dx=$

• A. $\sqrt{2}Si{n}^{-1}\left(\sin x+\cos x\right)+c$
• B. $\sqrt{2}{\mathrm{Cos}}^{-1}\left(\sin x+\cos x\right)+c$
• C. $\sqrt{2}{\mathrm{Cos}}^{-1}\left(\sin x-\cos x\right)+c$
• D. $\sqrt{2}Si{n}^{-1}\left(\sin x-\cos x\right)+c$

Answer 1

Answer: (D)

$\sqrt{2}Si{n}^{-1}\left(\sin x-\cos x\right)+c$

Solution-

$\int \sqrt{tanx}+\sqrt{cotx}dx=\int \frac{\sqrt{sinx}}{\sqrt{cosx}}+\frac{\sqrt{cosx}}{\sqrt{sinx}}dx$

$\int \frac{sinx+cosx}{\sqrt{sinxcosx}}dx=\sqrt{2}\int \frac{sinx+cosx}{\sqrt{2sinxcosx}}dx$

$=\sqrt{2}\int \frac{sinx+cosx}{\sqrt{2sinxcosx+1-1}}dx=\sqrt{2}\int \frac{sinx+cosx}{\sqrt{1-(sinx-cosx{)}^2}}dx$

put $sinx-cosx=t$ $(sinx+cosx)=dt$

$=\sqrt{2}\int \frac{dt}{\sqrt{1-t^2}}=\sqrt{2}si{n}^{-1}t+c$

$=\sqrt{2}si{n}^{-1}(sinx-cosx)+c$

D is correct.