Question

can someone solve int(e^x/(cos(x)))

Answer 1

Are you familiar with integration by parts?

Answer 2

yes ... but i am not able to solve using that ...

Answer 3

\[\int\limits_{?}^{?} udv = uv - \int\limits_{?}^{?} vdu\]

What should we choose as our u and dv?

Answer 4

may be
v=e^x
and
u=cos(x)

Answer 5

Sure. Carry out integration by parts once. What do you end up with?
(You should have an integral remaining on the right hand side that seems just as difficult to solve as this one. Leave it be for now)

Answer 6

Still there?

You should end up with

\[\int\limits\limits_{?}^{?}e ^{x}\cos x = e ^{x}\sin x - \int\limits\limits_{?}^{?}e ^{x}\sin x\]

You would then perform integration by parts on the last term again, which will give you:

\[\int\limits\limits_{?}^{?}e ^{x}\cos x = e ^{x}\sin x - (-e ^{x}\cos x + \int\limits_{?}^{?}e ^{x}\cos x )\]

Distribute the negateive, and move the integral term to the right hand side. Divide both sides by 2.

Answer 7

I thought he was trying to integrate
\[\int\limits_{}e^x/cosx dx\]

Answer 8

...Guess that'll teach me to read the problem carefully. One moment to do the correct work.

Answer 9

the problem is
\[\int{\frac{e^x}{\cos{x}}}\]

Answer 10

The actual problem I have is little more complex

\[\int{\frac{e^{ax}}{\sqrt{b+c\cos{dx}}}}\]

Answer 11

is the dx in the square root?

Answer 12

I am sorry ...
\[\int{\frac{e^{ax}}{\sqrt{b+c\cos{(x)}}}dx}\]

Answer 13

What level of math is this for? I'm having difficulty finding a way to make this clean...

Answer 14

I am working here to solve an model and got struck with this equation. It should be level of some master's engineering course.

Answer 15

Well, I went ahead and used some software to examine the two deceptively simple looking integrals. The answer to the first involves a hypergeometric function, and it couldn't solve the second.

Sorry, I don't think I'll be of much use to you on this.

Answer 16

Well.... thats explains why I'm stummped

Answer 17

\[\int\limits_{a}^{b}e^x/\cos x\]
let u=e^x, du=e^x dx, dx=du/e^x

so you have \[\int\limits_{a}^{b}du/cosx =\int\limits_{a}^{b} \sec x = \int\limits_{a}^{b}\ln \sec x + \tan x + C\]
Hope this helps

Answer 18

you cant do that because you have du/x, which means you have the change in u=du over cos(x), it has to either has to be du/u or dx/x or in the case of trig substitution dtheta/theta....... because of that your final answer lacks e^x and doesn't make sense