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appleduardo
whats the integral of sin x^(3) 3x^(2)?
1/36 (-81 (x^2-2) cos(x)+(9 x^2-2) cos(3 x)-6 x (sin(3 x)-27 sin(x)))+C
i know i have to use u du = -cos u +c, but dont know how to do it step by step :/
LEt x^3 = u see if it works ?
-cos(x)^(3) + C is that correct?
but what happened with du (which equals to 3x^2 ) ?
make the function as 3x^2* sinx^3
do u know trig identities of sin^2 X??
separate sin^3x into Sin^2x Sinx
its now become 3Integral X^2(sinX)(1/2(1-cos2x)dx
convert that into 3/2 int x^2(sinx)(1-cos2x)dx
Use the trigonometric identity sin(a) cos(b) 3/2 integral x^2 sin(x)dx - 3/4 integral x^2 (sin(3 x)-sin(x)) dx
its become 3/2 integral x^2 sin(x) dx - 3/4integral (x^2 sin(3 x)-x^2 sin(x)) dx
Integrate the sum term by term and factor out constants: = 3/4 integral x^2 sin(x) dx-3/4 integral x^2 sin(3 x) dx+3/2 integral x^2 sin(x) dx
im just gonna write it out what i do next.. just copy in out then u'll see
1/4 x^2 cos(3 x)+3/2 integral x^2 sin(x) dx+3/4 integral x^2 sin(x) dx-1/2 integral x cos(3 x) dx
= 1/4 x^2 cos(3 x)+3/2 integral x^2 sin(x) dx+3/4 integral x^2 sin(x) dx-1/6 x sin(3 x)+1/6 integral sin(3 x) dx
= 1/18 integral sin(u) du+1/4 x^2 cos(3 x)+3/2 integral x^2 sin(x) dx+3/4 integral x^2 sin(x) dx-1/6 x sin(3 x)
1/18 integral sin(u) du-3/4 x^2 cos(x)+1/4 x^2 cos(3 x)+3/2 integral x^2 sin(x) dx-1/6 x sin(3 x)+3/2 integral x cos(x) dx
1/18 integral sin(u) du-3/4 x^2 cos(x)+1/4 x^2 cos(3 x)+3/2 integral x^2 sin(x) dx+3/2 x sin(x)-1/6 x sin(3 x)-3/2 integral sin(x) dx
= -(cos(u))/18-1/4 (3 x^2 cos(x))+1/4 x^2 cos(3 x)+3/2 integral x^2 sin(x) dx+3/2 x sin(x)-1/6 x sin(3 x)-3/2 integral sin(x) dx = -(cos(u))/18-3/4 x^2 cos(x)+1/4 x^2 cos(3 x)+3/2 integral x^2 sin(x) dx+3/2 x sin(x)-1/6 x sin(3 x)+(3 cos(x))/2 = -(cos(u))/18-9/4 x^2 cos(x)+1/4 x^2 cos(3 x)+3/2 x sin(x)-1/6 x sin(3 x)+(3 cos(x))/2+3 integral x cos(x) dx = -(cos(u))/18-9/4 x^2 cos(x)+1/4 x^2 cos(3 x)+9/2 x sin(x)-1/6 x sin(3 x)+(3 cos(x))/2-3 integral sin(x) dx The integral of sin(x) is -cos(x): = -(cos(u))/18-9/4 x^2 cos(x)+1/4 x^2 cos(3 x)+9/2 x sin(x)-1/6 x sin(3 x)+(9 cos(x))/2+constant Substitute back for u = 3 x: = -9/4 x^2 cos(x)+1/4 x^2 cos(3 x)+9/2 x sin(x)-1/6 x sin(3 x)+(9 cos(x))/2-1/18 cos(3 x)+constant Which is equal to: Answer: | | = 1/36 (-81 (x^2-2) cos(x)+(9 x^2-2) cos(3 x)-6 x (sin(3 x)-27 sin(x)))+constant
I happen to did this before so im pretty sure its right .. but good luck,, check it w ur teacher , but if it wrong then please dont kill me ^^
oh my gosh! what a long integral! thank you so much for help me out!, and be sure i wont kill u if its wrong :P :D ^^
hmm, I did not read what you wrote but certainly no need for all that. after letting x^3 = u, => 3x^2 dx= du so integral becomes (sin u du) => -cos u + C =>-cos(x^3) + C as simple as that.