# int sin^4(6theta) d theta Find the indefinite integral

## Expert Answers

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

thenint 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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