Student Question

`int cos(5theta)cos(3theta) d theta` Find the indefinite integral

Expert Answers

An illustration of the letter 'A' in a speech bubbles

Indefinite integrals are written in the form of `int f(x) dx = F(x) +C`

 where:` f(x)` as the integrand

           `F(x)` as the anti-derivative function 

           `C`  as the arbitrary constant known as constant of integration

For the given problem `int cos(5theta)cos(3theta) d theta` has an integrand in a form of a trigonometric function. To evaluate this, we apply the identity:

`cos(X)cos(Y) =[cos(X+Y) +cos(X-Y)]/2`

The integral becomes:

`int cos(5theta)cos(3theta) d theta = int[cos(5theta+3theta) + cos(5theta -3theta)]/2 d theta`

 Apply the basic properties of integration:` int c*f(x) dx= c int f(x) dx` .

`int[cos(5theta+3theta) + cos(5theta -3theta)]/2d theta = 1/2int[cos(5theta+3theta) + cos(5theta -3theta)] d theta`

 Apply the basic integration property: `int (u+v) dx = int (u) dx + int (v) dx` .

`1/2 *[int cos(5theta +3theta)d theta+cos(5theta -3theta)d theta]`

Then apply u-substitution to be able to apply integration formula for cosine function:` int cos(u) du= sin(u) +C` .

For the integral: `int cos(5theta +3theta)d theta` , we let ` u =5theta +3theta =8theta` then `du= 8 d theta` or `(du)/8 =d theta` .

`int cos(5theta +3theta)d theta=int cos(8theta)d theta`

                                 `=intcos(u) *(du)/8`

                                 `= 1/8 int cos(u)du`

                                 `= 1/8 sin(u) +C`

Plug-in `u =8theta` on `1/8 sin(u) +C` , we get:

`int cos(5theta +3theta)d theta=1/8 sin(8theta) +C`

 For the integral: `intcos(5theta -3theta)d theta` , we let `u =5theta -3theta =2theta` then `du= 2 d theta` or `(du)/2 =d theta` .

`int cos(5theta -3theta)d theta = intcos(2theta) d theta`

                                `=intcos(u) *(du)/2`

                                `= 1/2 int cos(u)du`

                                `= 1/2 sin(u) +C`

Plug-in `u =2 theta` on `1/2 sin(u) +C` , we get:

`intcos(5theta -3theta)d theta =1/2 sin(2theta) +C`

Combing the results, we get the indefinite integral as:

`int cos(5theta)cos(3theta)d theta = 1/2*[1/8 sin(8theta) +1/2 sin(2theta)] +C`

or   `1/16 sin(8theta) +1/4 sin(2theta) +C`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial