What is the Integration of e^x*([1+sin x]/[1+cos x])dx?

Question

The answer is tan(x/2)+c but how it is coming?

Answer 1

This is a difficult integral. You should be able to do it by multiplying the top and bottom of the integrand by (1-cos x). This would turn the denominator into sin^2 x and after simplifying the integrand would turn into:

e^x ( csc^2 x + csc x - (csc x)(cot x) - cot x ).

This is messy but doable. However, the answer is definitely NOT tan(x/2)+C.

The derivative of tan(x/2)+C is (1/2)sec^2(x/2) which doesn't have an e^x in it ...

Answer 2

Your answer is incorrect... should be e^x * tan(x/2) + c.

Let u = x/2, and rewrite sin x and cos x as

[1] ... sin x = 2 sin u cos u
[2] ... cos x = 2(cos u)^2 - 1

The fraction [1 + sin x]/[1 + cos x] now becomes

[3] ... [1 + 2 sin u cos u] / [2(cos u)^2]

which can be reduced to

[4] ... 1/[2(cos u)^2] + tan u

This shows that the original formula

[5] ... e^x * ([1+sin x]/[1 + cos x]) =
[6] ... e^x/[2(cos u)^2] + e^x tan u

This is equal to

[7] ... e^x . d(tan u)/dx + e^x . tan u

and, using product rule, equal to the derivative of

[8] ... e^x . tan u

For the integral we find therefore

[9] ... e^x . tan (x/2) + c

===

Okay, I have to be honest... I guessed you missed a factor e^x in your answer, checked it numerically with my computer, and then worked my way back from there. This integral is tricky and you should not be hard on yourself if you couldn't find it :)

Answer 3

Someone will know this answer, I on the other hand think there is a flaw in the math, and any hypothesis is moot!

Answer 4

Work backwards, take the d/dx and then jsut fiddle with identities and stuff.

Answer 5

you had too integrate by parts for a lonng while and maybe use some trigonometrics

Answer 6

who gives a crap?