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# ntegral of sin^3(6x)cos^4(3x) dx ?

between limits o to 60 degrees

### 4 Answers

- θ βяιαη θLv 78 years agoFavorite Answer
I am assuming you want to compute the integral from 0 to π/3 (as sin^3(6x)cos^4(3x)'s anti-derivative is different when considering degrees, and you probably don't want this).

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Using the double-angle formula for sine to knock down the 6x argument to 3x gives:

sin^3(6x)cos^4(3x) = [2sin(3x)cos(3x)]^3*cos^4(3x) = 8sin^3(3x)cos^7(3x).

So:

∫ sin^3(6x)cos^4(3x) dx = ∫ 8sin^3(3x)cos^7(3x) dx = 8 ∫ sin^3(3x)cos^7(3x) dx.

Then, re-writing cos^7(3x) as cos(3x)cos^6(3x) = cos(3x)[1 - sin^2(3x)]^3 gives:

8 ∫ sin^3(3x)cos^7(3x) dx = 8 ∫ sin^3(3x)[1 - sin^2(3x)]^3 [cos(3x) dx].

Applying the substitution u = sin(3x) ==> du = 3sin(3x) dx and cos(3x) dx = (1/3)du:

8 ∫ sin^3(3x)[1 - sin^2(3x)]^3 [cos(3x) dx] = 8/3 ∫ u^3(1 - u^2)^3 du

= 8/3 ∫ u^3(1 - 3u^2 + 3u^4 - u^6) du, by binomial expansion

= 8/3 ∫ (u^3 - 3u^5 + 3u^7 - u^9) du, by expanding

= (2/3)u^4 - (4/3)u^6 + u^8 - (4/15)u^10 + C, by integrating term-by-term.

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Thus, back-substituting u = sin(3x) yields the corresponding indefinite integral to be:

(2/3)sin^4(3x) - (4/3)sin^6(3x) + sin^8(3x) - (4/15)sin^10(3x) + C.

Therefore, applying the limits yields that the required integral equals:

[(2/3)sin^4(3x) - (4/3)sin^6(3x) + sin^8(3x) - (4/15)sin^10(3x)] (evaluated from x=0 to π/3)

= 0, since each term evaluates to 0 at both x = 0 and x = π/3.

- ANONYMOUSLv 68 years ago
First of all, Let u=3X Then du=3dX and your integral becomes 1/3 integral(sin^3(u)cos^4(u)du

=1/3integral{sinusin^2(u)cos^4(u)du}=1/3integral{sinu[1-cos^2(u)]cos^4(u)du}. Let v=cosu, dv= -sinu

so your integral becomes 1/3integral{-v^4(1-v^2)dv}=1/3integral{v^6-v^4dv}=v^7/21 +v^5/15

=cos^7(3x)/21 +cos^5(3x)/15. I will let you evaluate the rest.

- Anonymous8 years ago
Hello Mohammad,

This integral can be greatly simplified using a change of variables so that we may apply the odd-function rule. I have demonstrated this at one of the math forums I help to moderate, so that I may use LaTeX to give you an easy to read explanation:

- Anonymous8 years ago
Hmm its kinda hard questions but answer is;

Write sin^3(6x)cos^4(3x) = 1/64 (14sin(6x) + 8sin(12x) -3sin(18x) - 4sin(24x)-sin(30x))

and integrate it 1/64 ∫ (14sin(6x) + 8sin(12x) -3sin(18x) - 4sin(24x)-sin(30x)) dx

integrate the sum, term by term and factor out constants:

= 7/32 ∫ sin(6x)dx + 1/8 ∫ sin(12x)dx - 3/64 ∫ sin(18x)dx - 1/16 ∫ sin(24x)dx - 1/64 ∫ sin(30x)dx

For the integrand sin(6x),substitute u= 6x and du=6dx

= 7/192 ∫ sin(u)du + 1/8 ∫ sin(12x)dx - 3/64 ∫ sin(18x)dx - 1/16 ∫ sin(24x)dx - 1/64 ∫ sin(30x)dx

For the integrand sin(12x), substitute s=12x and ds=12dx

=1/96∫ sin(s)ds + 7/192 ∫ sin(u)du - 3/64 ∫ sin(18x)dx - 1/16 ∫ sin(24x)dx - 1/64 ∫ sin(30x)dx

For the integrand sin(18x), substitute p=18x and dp = 18dx:

= -1/384 ∫ sin(p)dp + 1/96∫ sin(s)ds+ 7/192 ∫ sin(u)du- 1/16 ∫ sin(24x)dx - 1/64 ∫ sin(30x)dx

For the integrand sin(24x),substitute w=24x and dw=24dx;

= -1/384 ∫ sin(p)dp + 1/96∫ sin(s)ds+ 7/192 ∫ sin(u)du - 1/384 ∫sin(w)dw -1/64 ∫ sin(30x)dx

For the integrand sin(30x), substitute v=30x and dv=30dx;

-1/384 ∫ sin(p)dp + 1/96∫ sin(s)ds+ 7/192 ∫ sin(u)du - 1/384 ∫sin(w)dw - 1/1920∫sin(v)dv

Integral result is;

cos(p)/384 - cos(s)/96 - 7cos(u)/192 + cos(v)/1920 + cos(w)/384 + c

Substitute back and solution is;

1/1920(-70cos(6x)-20cos(12x)+5cos(18x)+5cos(24x)+cos(30x)) + c