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

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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)...

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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`

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