Vi Nguyen
May 1, 2009, 09:11 PM
Can anyone help me in finding the antiderivative of:

Sec(x)e^(tan(x))

Antidifferentiation is so confusing!

galactus
May 2, 2009, 05:26 AM
\int sec^{2}(x)e^{tan(x)}dx

Actually, this one is not that bad if you see the sub to make.

Let u=tan(x), \;\ du=sec^{2}(x)dx

Make the subs and it's easy.

Vi Nguyen
May 2, 2009, 10:42 AM
Does this mean this mean that I find the integral of e^(tan(X)), then is the answer e^(tan(x))/tan(x) +C ?

\int sec^{2}(x)e^{tan(x)}dx

Actually, this one is not that bad if you see the sub to make.

Let u=tan(x), \;\ du=sec^{2}(x)dx

Make the subs and it's easy.

galactus
May 2, 2009, 10:49 AM
No, just make the substitution. Have you learned you substitution?.

If we let u=tan(x), \;\ du=sec^{2}(x)dx, it whittles down to

\int e^{\overbrace{u}^{\text{tan(x)}}}\underbrace{du}_{ \text{sec^2(x)dx}}

See why?. The sec^{2}(x)dx is taken care of by the du and the tan(x) is you as

The power in e.

Now, integrate and resub. e^{u} is the easiest to integrate because it stays the same.

Anti-differentiation is not that confusing.Just think of it as the opposite of differentiation.

Once you find the anti-derivative, differentiate and you should get back to the original, e^{tan(x)}sec^{2}(x)

Vi Nguyen
May 2, 2009, 07:00 PM

No, just make the substitution. Have you learned you substitution?.

If we let u=tan(x), \;\ du=sec^{2}(x)dx, it whittles down to

\int e^{\overbrace{u}^{\text{tan(x)}}}\underbrace{du}_{ \text{sec^2(x)dx}}

See why?. The sec^{2}(x)dx is taken care of by the du and the tan(x) is you as

The power in e.

Now, integrate and resub. e^{u} is the easiest to integrate because it stays the same.

Anti-differentiation is not that confusing.Just think of it as the opposite of differentiation.

Once you find the anti-derivative, differentiate and you should get back to the original, e^{tan(x)}sec^{2}(x)