Question-212239

Question Number 212239 by Spillover last updated on 07/Oct/24

Answered by Ghisom last updated on 08/Oct/24

\int \frac{e^{- x / 2} \sqrt{1 - \sin x}}{1 + \cos x} d x =

[t = \tan \frac{x}{2}]

= \int \frac{e^{- \arctan t} (t - 1)}{\sqrt{t^{2} + 1}} d t =

= \int \frac{e^{- \arctan t} t}{\sqrt{t^{2} + 1}} d t - \int \frac{e^{- \arctan t}}{\sqrt{t^{2} + 1}} d t =

[by parts]

= e^{- \arctan t} \sqrt{t^{2} + 1} + \int \frac{e^{- \arctan t}}{\sqrt{t^{2} + 1}} d t - \int \frac{e^{- \arctan t}}{\sqrt{t^{2} + 1}} d t =

= e^{- \arctan t} \sqrt{t^{2} + 1} =

= \frac{e^{- x / 2}}{\cos \frac{x}{2}} + C

Commented by Frix last updated on 08/Oct/24

Nice!

Commented by Spillover last updated on 08/Oct/24

g r e a t . t h a n k s

Commented by MathematicalUser2357 last updated on 12/Oct/24

No, the real answer is \frac{e^{- x / 2} \sqrt{1 - \sin x} \sec^{2} \frac{x}{2}}{\tan \frac{x}{2} + 1} + C .

Commented by Ghisom last updated on 13/Oct/24

{it}^{'} s the same, just transform it to see

Answered by MathematicalUser2357 last updated on 12/Oct/24

\frac{e^{- x / 2} \sqrt{1 - \sin x} \sec^{2} \frac{x}{2}}{\tan \frac{x}{2} - 1} + C