# `int sin(theta)sin(3theta) d theta` Find the indefinite integral

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 sin(theta)sin(3theta) d theta`  or  `int sin(3theta)sin(theta) d theta`   has a integrand in a form of trigonometric function. To evaluate this, we apply the identity:

`sin(A)sin(B) =[-cos(A+B) +cos(A-B)]/2`

The integral becomes:

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

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

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

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

`1/2 *[int -cos(3theta+theta)d theta+cos(3theta -theta)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(3theta+theta)d theta` , we let` u =3theta +theta =4theta` then `du= 4 d theta` or `(du)/4 =d theta` .

`int -cos(3theta+theta)d theta=int -cos(4theta)d theta`

`=int -cos(u) *(du)/4`

`= -1/4 int cos(u)du`

`= -1/4 sin(u) +C`

Plug-in `u =4theta` on `-1/4 sin(u) +C` , we get:

`int -cos(3theta+theta)d theta= -1/4 sin(4theta) +C`

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

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

Combining the results, we get the indefinite integral as:

`intsin(theta)sin(3theta)d theta = 1/2*[ -1/4 sin(4theta) +1/2 sin(2theta)] +C`

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