Evaluate-tan-d- – Tinku Tara

Question Number 119 by novrya last updated on 25/Jan/15

E v a l u a t e \int \sqrt{t a n θ} d θ

Answered by rajabhay last updated on 06/Dec/14

\tan θ = t^{2}

\sec^{2} θ d θ = 2 t d t, \sec^{2} θ = 1 + t^{4}

d θ = \frac{2 t d t}{1 + t^{4}}

\int \sqrt{\tan θ} d θ = \int \frac{2 t^{2}}{1 + t^{4}} d t

= \int \frac{t^{2} + 1}{1 + t^{4}} d t + \int \frac{t^{2} - 1}{1 + t^{4}} d t

= \int \frac{1 + 1 / t^{2}}{t^{2} + 1 / t^{2}} d t + \int \frac{1 - 1 / t^{2}}{t^{2} + 1 / t^{2}}

= \int \frac{1 + 1 / t^{2}}{t^{2} + 1 / t^{2} - 2 + 2} d t + \int \frac{1 - 1 / t^{2}}{t^{2} + 1 / t^{2} + 2 - 2}

= \int \frac{1 + 1 / t^{2}}{{(t - 1 / t)}^{2} + 2} d t + \int \frac{1 - 1 / t^{2}}{{(t + 1 / t)}^{2} - 2}

t - 1 / t = u in 1 st intergral and t + 1 / t = v in 2 nd

(1 + 1 / t^{2}) d t = d u, (1 - 1 / t^{2}) d t = d v

So the integral now equals

\int \frac{d u}{u^{2} + 2} + \int \frac{d v}{v^{2} - 2}

= \frac{1}{\sqrt{2}} \tan^{- 1} \frac{u}{\sqrt{2}} + \frac{1}{2 \sqrt{2}} \ln ∣ \frac{v - \sqrt{2}}{v + \sqrt{2}} ∣ + C

= \frac{1}{\sqrt{2}} \tan^{- 1} \frac{1}{\sqrt{2}} (t - \frac{1}{t}) + \frac{1}{2 \sqrt{2}} \ln ∣ \frac{t + 1 / t - \sqrt{2}}{t + 1 / t + \sqrt{2}} ∣ + C

= \frac{1}{\sqrt{2}} \tan^{- 1} \frac{1}{\sqrt{2}} (\frac{t^{2} - 1}{t}) + \frac{1}{2 \sqrt{2}} \ln ∣ \frac{t^{2} + 1 - t \sqrt{2}}{t^{2} + 1 + t \sqrt{2}} ∣ + C

= \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\tan θ - 1}{\sqrt{2 \tan θ}}) + \frac{1}{2 \sqrt{2}} \ln ∣ \frac{\tan θ + 1 - \sqrt{2 \tan θ}}{\tan θ + 1 + \sqrt{2 \tan θ}} ∣ + C

Commented by malwaan last updated on 07/Sep/16

g r e e t

Commented by rajabhay last updated on 25/Sep/16

Thanks

Commented by peter frank last updated on 05/Jan/22

thanks