Ask Me Help Desk - How to find the antiderivative of trig

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How to find the antiderivative of trig

Can anyone help me in finding the antiderivative of:

sec²(x)e^(tan(x))

Antidifferentiation is so confusing!!

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

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.

Does this mean this mean that I find the integral of e^(tan(X)), then is the answer e^(tan(x))/tan(x) +C ?

Quote:

Originally Posted by galactus

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

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.

No, just make the substitution. Have you learned u 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 u 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, $e^{tan(x)}sec^{2}(x)$

Thanks you're a legend! ;p

Quote:

Originally Posted by galactus

No, just make the substitution. Have you learned u 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 u 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, $e^{tan(x)}sec^{2}(x)$

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