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
To evaluate the given problem `int sin^4(6theta) d theta` , we may apply u-substitution by letting: `u = 6theta` then `du = 6 d theta` or `(du)/6 = d theta` .
The integral becomes:
`int sin^4(6theta) d theta=int sin^4(u) * (du)/6`
Apply the basic properties of integration: `int c*f(x) dx= c int f(x) dx` .
`int sin^4(u) * (du)/6=1/6int sin^4(u)du` .
Apply the integration formula for sine function: `int sin^n(x) dx = -(cos(x)sin^(n-1)(x))/n+(n-1)/n int sin^(n-2)(x)dx` .
`1/6int sin^4(u)du=1/6[-(cos(u)sin^(4-1)(u))/4+(4-1)/4 int sin^(4-2)(u)du]` .
`=1/6[-(cos(u)sin^(3)(u))/4+3/4 int sin^(2)(u)du]`
For the integral `int sin^(2)(u)du` , we may apply trigonometric identity: `sin^2(x)= 1-cos(2x)/2 or 1/2 - cos(2x)/2.`
We get:
`int sin^(2)(u)du = int ( 1/2 - cos(2u)/2) du` .
Apply the basic integration property:`int (u-v) dx = int (u) dx - int (v) dx` .
`int ( 1/2 - cos(2u)/2) du=int ( 1/2) du - int cos(2u)/2 du`
`= 1/2u - 1/4sin(2u)+C`
or `u/2 - sin(2u)/4+C`
Note: From the table of integrals, we have `int cos(theta) d theta = sin(theta)+C.`
Let: `v = 2u` then `dv = 2du ` or` (dv)/2= du`
then`int cos(2x)/2 du =int cos(v)/2 * (dv)/2`
`= 1/4 sin(v)`
`= 1/4 sin(2u)`
Applying `int sin^(2)(u)du=u/2 - sin(2u)/4+C` , we get:
`1/6int sin^4(u)du=1/6[-(cos(u)sin^(3)(u))/4+3/4 int sin^(2)(u)du]`
`=1/6[-(cos(u)sin^(3)(u))/4+3/4 [u/2 - sin(2u)/4]]+C`
`=1/6[-(cos(u)sin^(3)(u))/4+(3u)/8 - (3sin(2u))/16]+C`
`=(-cos(u)sin^(3)(u))/24+(3u)/48 - (3sin(2u))/96+C`
Plug-in `u =6theta ` on `(-cos(u)sin^(3)(u))/24+(3u)/48 - (3sin(2u))/96+C` to find the indefinite integral as:
`int sin^4(6theta) d theta =(cos(6theta)sin^(3)(6theta))/24+(3*6theta)/48 - (3sin(2*6theta))/96+C`
`=(cos(6theta)sin^(3)(6theta))/24+(18theta)/48 - (3sin(12theta))/96+C`
`=(cos(6theta)sin^(3)(6theta))/24+(3theta)/8 - (sin(12theta))/32+C`
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