C-1-tan-2-x-1-sec-2-x-dx-

Question Number 166449 by cortano1 last updated on 20/Feb/22

C = \int \frac{1 - \tan^{2} x}{1 + \sec^{2} x} dx = ?

Commented by cortano1 last updated on 20/Feb/22

oo yes

Commented by cortano1 last updated on 20/Feb/22

Commented by MJS_new last updated on 20/Feb/22

u = \tan \frac{x}{2} \Leftrightarrow x = 2 \arctan u \Rightarrow \cos^{2} x = {(\frac{1 - u^{2}}{1 + u^{2}})}^{2}

Answered by MJS_new last updated on 20/Feb/22

\int \frac{1 - \tan^{2} x}{1 + \sec^{2} x} d x = 2 \int d x - 3 \int \frac{1}{1 + \cos^{2} x} d x =

[t = \tan x \to d x = \cos^{2} x d t]

= 2 x - 3 \int \frac{d t}{t^{2} + 2} =

= 2 x - \frac{3}{\sqrt{2}} \arctan \frac{t}{\sqrt{2}} =

= 2 x - \frac{3}{\sqrt{2}} \arctan \frac{\tan x}{\sqrt{2}} + C

Commented by peter frank last updated on 24/Feb/22

great