What's the integral of #int (tanx)*(e^x)dx#?

Question

Cameron Alexander · Accepted Answer

Some guidence:

This is a very interesting questions, but i didnt get to far finding an exact antiderivative solution, then i also tried an online integral calculator....

There is one way i can think of that can approximate a solutio:

Use the Macluaren series:

#y(x) = sum_(n=0) ^oo (y^(n)(0)* x^n)/(n!) #

Letting #y(x) = e^x * tanx #

#=> y'(x) = e^x ( tanx + sec^2 x ) #

#=> y''(x) = e^x (tanx + 2sec^2 x + 2sec^2x tanx )#
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Then evaluating each of these at #x = 0# and using the maclauren series, then integrating using power rule:

#=> int # # sum_(n=0) ^oo (y^(n)(0)* x^n)/(n!)# # dx #

Hope this was a good step in the right direction to an approximat answer!

Now here is an interesting question!:

#int# # e^x sinx# # dx # = ??

Give it a go!

Samantha Adkison · Answer

undefined The integral of $ \int \tan(x) \cdot e^x \, dx $ can be found using integration by parts.

Let $ u = \tan(x) $ and $ dv = e^x \, dx $.

Then, $ du = \sec^2(x) \, dx $ and $ v = e^x $.

Applying integration by parts formula:
$$ \int u \, dv = uv - \int v \, du $$

$$ \int \tan(x) \cdot e^x \, dx = \tan(x) \cdot e^x - \int e^x \cdot \sec^2(x) \, dx $$

The integral $ \int e^x \cdot \sec^2(x) \, dx $ is straightforward to solve, as it's a standard integral.

$$ \int e^x \cdot \sec^2(x) \, dx = e^x \cdot \tan(x) - \int e^x \cdot \tan(x) \, dx $$

Adding $ \int e^x \cdot \tan(x) \, dx $ to both sides gives:
$$ 2 \int e^x \cdot \tan(x) \, dx = e^x \cdot \sec^2(x) + e^x \cdot \tan(x) $$

Dividing both sides by 2 gives the result:
$$ \int e^x \cdot \tan(x) \, dx = \frac{1}{2} e^x \cdot \sec^2(x) + \frac{1}{2} e^x \cdot \tan(x) + C $$

Where $ C $ is the constant of integration.