dx-sinx-cosx-1-

Question Number 160432 by mkam last updated on 29/Nov/21

\int \frac{d x}{s i n x + c o s x + 1}

Answered by Ar Brandon last updated on 29/Nov/21

I = \int \frac{d x}{\sin x + \cos x + 1}

= \int \frac{d x}{2 \sin (\frac{x}{2}) \cos (\frac{x}{2}) + {2 \cos}^{2} \frac{x}{2}}

= \frac{1}{2} \int \frac{\sec^{2} (\frac{x}{2})}{\tan (\frac{x}{2}) + 1} d x = \int \frac{d (\tan \frac{x}{2})}{\tan \frac{x}{2} + 1}

= \ln ∣ \tan \frac{x}{2} + 1 ∣ + C

Commented by MJS_new last updated on 29/Nov/21

I had solved it, too .

but we only get “ ? ? ? ? ?^{″} from this member

if we {don}^{'} t solve his questions fast enough

so I decided to delete my solution .

Commented by Ar Brandon last updated on 29/Nov/21

Well \dots what can we do, Sir ? Ignoring

and reacting when we feel like doing so,

in my opinion is the best way . Btw I {don}^{'} t

feel pressurized by any “ ? ? ?^{″} .

Commented by MJS_new last updated on 29/Nov/21

me neither . I just decided not to answer any

more .

Commented by mr W last updated on 30/Nov/21

“ ? ? ? ?^{″} m e a n s n o t h i n g b u t “ h e l p! h e l p!,

b e q u i c k! w h y s t i l l n o h e l p ?^{″} . i f i t i s

o n l y t h i s, {i t}^{'} s s t i l l o k t o m e . b u t t h e

p r o b l e m i s w h e n y o u r e a l l y h e l p, n o

r e s p o n s e c o m e s b a c k . a m a n c a n c r y

f o r h e l p, b u t i s n o t a b l e t o s a y t h a n k s

t o t h e h e l p e r s .

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