please help me with this integral question!?

Question

What is the integral of this function:f(x)=sin^4(x)cos^6(x)?

Answer 1

f(x) = Integral ( sin^4(x) cos^6(x) dx)

Express cos^6(x) as cos^4(x) cos^2(x).

f(x) = Integral ( sin^4(x) cos^4(x) cos^2(x) dx )

Express sin^4(x) cos^4(x) as [ sin(x) cos(x) ]^4.

f(x) = Integral ( [ sin(x)cos(x) ]^4 cos^2(x) dx )

The half angle identity states that

sin(2x) = 2sin(x) cos(x), so
(1/2)sin(2x) = sin(x) cos(x).
Replace the term which is being put to the power of 4 accordingly.

f(x) = Integral ( (1/2)sin(2x) ]^4 cos^2(x) dx )

Which simplifies as

f(x) = Integral ( 1/16 sin^4(2x) cos^2(x) dx )

Factor the (1/16) from the integral.

f(x) = (1/16) Integral ( sin^4(2x) cos^2(x) dx )

Use the half angle identity cos^2(x) = (1/2) (1 + cos(2x)).

f(x) = (1/16) Integral ( sin^4(2x) (1/2) (1 + cos(2x)) dx )

Factor the (1/2) from the integral.

f(x) = (1/32) Integral ( sin^4(2x) (1 + cos(2x)) dx )

Expand.

f(x) = (1/32) Integral ( [ sin^4(2x) + sin^4(2x) cos(2x) ] dx )

Split into two integrals.

f(x) = (1/32) [ Integral ( sin^4(2x) dx ) + Integral ( sin^4(2x) cos(2x) dx ) ]

Distribute the 1/32.

f(x) = (1/32) Integral ( sin^4(2x) dx ) + (1/32) Integral ( sin^4(2x) cos(2x) dx )

We can solve the second integral using substitution, and the first integral using half angle identities. Let's solve them one at a time.

Let a(x) = (1/32) Integral ( sin^4(2x) dx )
b(x) = (1/32) Integral ( sin^4(2x) cos(2x) dx )

(To keep a record of things, our answer should be
f(x) = a(x) + b(x) + C)

1) a(x) = (1/32) Integral ( sin^4(2x) dx )

Express sin^4(2x) as [ sin^2(2x) ]^2

a(x) = (1/32) Integral ( [ sin^2(2x) ]^2 dx )

Use the half angle identity
sin^2(2x) = (1/2) (1 - cos(2*2x)).

a(x) = (1/32) Integral ( [(1/2) (1 - cos(2*2x) ]^2 dx )

a(x) = (1/32) Integral ( [(1/2) (1 - cos(4x)) ]^2 dx )

a(x) = (1/32) Integral ( (1/4) (1 - cos(4x))^2 dx )

a(x) = (1/128) Integral ( (1 - cos(4x))^2 dx )

a(x) = (1/128) Integral ( (1 - 2cos(4x) + cos^2(4x)) dx )

a(x) = (1/128) (x - 2sin(4x)(1/4) + Integral ( cos^2(4x) dx ) )

a(x) = (1/128)x - (1/64)sin(4x) + (1/128) Integral ( cos^2(4x) dx)

a(x) = (1/128)x - (1/64)sin(4x) + (1/128) Integral ( (1/2)(1 + cos(8x)) dx )

a(x) = (1/128)x - (1/64)sin(4x) + (1/256) Integral ( (1 + cos(8x)) dx )

a(x) = (1/128)x - (1/64)sin(4x) + (1/256) (x + (1/8)sin(8x))

a(x) = (1/128)x - (1/64)sin(4x) + (1/256)x + (1/2048)sin(8x)

a(x) = (3/256)x - (1/64)sin(4x) + (1/2048)sin(8x)

Tall order there. Let's now solve for the second half, b(x).

2) b(x) = (1/32) Integral ( sin^4(2x) cos(2x) dx )

Solve by substitution.

Let u = sin(2x).
du = cos(2x) (2) dx
(1/2) du = cos(2x) dx

b(x) = (1/32) Integral ( u^4 (1/2) du )
b(x) = (1/64) Integral ( u^4 du )
b(x) = (1/64) (1/5)u^5
b(x) = (1/320) u^5
b(x) = (1/320) sin^5(2x)

Combine the two halves together. Don't forget to add a constant.

f(x) = a(x) + b(x) + C.
f(x) = (3/256)x - (1/64)sin(4x) + (1/2048)sin(8x) + (1/320) sin^5(2x) + C

Answer 2

Integral sin^m(x)cos^n(x) dx =
(-sin^(m-1) x cos ^(n-1) x )/(m+n)
+ (m-1)/(m+n) *intergral sin^(m-2) x cos^n x dx

Repeat as necessary.

Answer 3

sin^4(x)cos^6(x) is the same as
sin(x)^4cos(x)^6 which integrates to :-

wait for it........


-cos(x)^7[(sin(x)^3)/10 + (3sin(x))/80] + (sin(x)cos(x)^5)/160 + (sin(x)cos(x)^3)/128 + (3sin(x)cos(x))/256 + 3(x)/256 + k

its so hard to write formulae in Y!Answers.

it reads as follows (hope this helps)

minus cosx to the power of 7 multiplied by (open bracket) sinx cubed over 10 plus 3 sinx over 80 (closed bracket) plus sinx times cosx to the power of 5 all over 160 plus sinx times cosx cubed all over 128 plus 3sinx times cosx all over 256 plus 3x over 256 plus k