# `int sin5xsin4x dx` 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(5theta)sin(4theta) d theta`  has an integrand in the form of a trigonometric function. To evaluate this, we apply the identity:

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

The integral becomes:

`intsin(5theta)sin(4theta)d theta= int[-cos(5theta+4theta) + cos(5theta -4theta)]/2 d theta`

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

`int[-cos(5theta+4theta) + cos(5theta -4theta)]/2 d theta= 1/2int[-cos(5theta+4theta) +cos(5theta -4theta)] d theta`

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

`1/2 *[int-cos(5theta+4theta)d theta+cos(5theta -4theta)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+4theta)`   d theta, we let` u =5theta +4theta =9theta` then `du= 9 d theta` or `(du)/9 =d theta` .

`int -cos(5theta+4theta)d theta=int -cos(9theta)d theta`

`=int -cos(u) *(du)/9`

`= -1/9 int cos(u)du`

`= -1/9 sin(u) +C`

Plug-in `u =9theta` on `-1/9 sin(u) +C` , we get:

`int-cos(5theta+4theta)d theta= -1/9 sin(9theta) +C`

For the integral: `intcos(5theta -4theta)d theta` , we let `u =5theta -4theta =theta` then `du= d theta` .

`intcos(5theta -4theta)d theta = intcos(theta) d theta`

`=intcos(u) *(du)`

`=sin(u) +C`

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

`intcos(5theta -4theta)d theta = sin(theta) +C`

Combining the results, we get the indefinite integral as:

`intsin(5theta)sin(4theta)d theta = 1/2*[ -1/9 sin(9theta) + sin(theta)] +C`

or  `- 1/18 sin(4theta) +1/2 sin(theta) +C`

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