find-2-tan-2-t-dt-

Question Number 40830 by math khazana by abdo last updated on 28/Jul/18

f i n d \int \sqrt{2 + {t a n}^{2} t} d t

Commented by maxmathsup by imad last updated on 31/Jul/18

l e t I = \int \sqrt{2 + {t a n}^{2} t} d t c h a n g e m e n t t a n t = \sqrt{2} s h (x) \Rightarrow t = a r c t a n (\sqrt{2} s h x)

\Rightarrow d t = \frac{\sqrt{2} c h (x)}{1 + 2 {s h}^{2} x} d x

I = \int \sqrt{2 + 2 {s h}^{2} x} \frac{\sqrt{2} c h (x)}{1 + 2 {s h}^{2} x} d x = \int \frac{2 {c h}^{2} x}{1 + 2 {s h}^{2}} d x

= \int \frac{1 + c h (2 x)}{1 + c h (2 x) - 1} d x = \int \frac{c h (2 x) + 1}{c h (2 x)} d x = x + \int \frac{d x}{c h (2 x)} b u t

\int \frac{d x}{c h (2 x)} = \int \frac{2 d x}{e^{2 x} + e^{- 2 x}} =_{e^{x} = t} \int \frac{2}{t^{2} + t^{- 2}} \frac{d t}{t}

= \int \frac{2 d t}{t^{3} + \frac{1}{t}} = \int \frac{2 t d t}{t^{4} + 1} =_{t^{2} = u} \int \frac{d u}{1 + u^{2}} = a r c t a n (u) + c

= a r c t a n (t^{2}) + c = a c t a n (e^{2 x}) + c b u t x = a r g s h (\frac{t a n t}{\sqrt{2}})

= l n (\frac{t a n t}{\sqrt{2}} + \sqrt{1 + \frac{{t a n}^{2} t}{2})})

I = a r g s h (\frac{t a n t}{\sqrt{2}}) + a r c t a n ({\frac{t a n t}{\sqrt{2}} + \sqrt{1 + \frac{{t a n}^{2} t}{2}}}^{2}) + c