How do you integrate #int e^x cos ^2 x dx # using integration by parts?

Question

How do you integrate #int e^x  cos ^2   x    dx  # using integration by parts?

Ezekiel Alexander · Accepted Answer

#1/10 e^x cos 2x + 1/5 e^x sin 2x + 1/2 e^x + C#

First, consider the identity #cos 2x = 2cos^2 x - 1#. Use this identity to transform the integral

#int (e^x cos^2 x) dx#

into the integral

#int (e^x (cos 2x + 1))/2 dx = 1/2 int e^x cos 2x dx + 1/2 int e^x dx#

Finding #int e^x dx# is straightforward i.e. #int e^x dx = e^x + C#

We use integration by parts to find #int e^x cos 2x dx#. By LIATE, we integrate #e^x# and differentiate #cos 2x#:

#int (e^x cos 2x) dx = e^x cos 2x - int e^x (–2 sin 2x) dx = e^x cos 2x + 2 int (e^x sin 2x) dx = e^x cos 2x + 2[(e^x sin 2x)-int e^x (2 cos 2x) dx] = e^x cos 2x + 2e^x sin 2x-4 int (e^x cos 2x) dx#

Thus,

#int (e^x cos 2x) dx = e^x cos 2x + 2e^x sin 2x-4 int (e^x cos 2x) dx# #5int (e^x cos 2x) dx = e^x cos 2x + 2e^x sin 2x dx# #int (e^x cos 2x) dx = 1/5 e^x cos 2x + 2/5 e^x sin 2x dx#

Therefore,

#int (e^x (cos 2x + 1))/2 dx = 1/2 (1/5 e^x cos 2x + 2/5 e^x sin 2x) + 1/2 e^x + C = 1/10 e^x cos 2x + 1/5 e^x sin 2x + 1/2 e^x + C#

Samantha Adkison · Answer

undefined To integrate $ \int e^x \cos^2(x) \, dx $ using integration by parts, you can follow these steps:

1. Choose $ u $ and $ dv $.
2. Calculate $ du $ and $ v $.
3. Apply the integration by parts formula: $ \int u \, dv = uv - \int v \, du $.

Let's solve it:

Choose $ u = \cos^2(x) $ and $ dv = e^x \, dx $.
Then, calculate $ du $ and $ v $:
$ du = -2\cos(x) \sin(x) \, dx $
$ v = e^x $

Now, apply the integration by parts formula:
$$ \int e^x \cos^2(x) \, dx = \cos^2(x) \cdot e^x - \int e^x \cdot (-2\cos(x) \sin(x)) \, dx $$

This integral can be further simplified, but since it's a new integral, you can apply integration by parts again or use other methods to solve it.