Thread: intergration

antidifferentiation of

(x^3)/squareroot((3x^4)+5) dx

i know the formula 1/square root(1 - x^2) but that doesnt help me.


can some one please check this one for me

antidifferentiation

sin^2xcos^2x dx

i got 1/3sinx^3 - 1/2sinx^4 + 1/5sin^5 + C

not very confident with this answer

also could some one check the antidifferentiation of

sin3xcos4x dx

i got 1/4sinx - 1/14sinx + C

thanks in advance

Originally Posted by helmszee

antidifferentiation of

(x^3)/squareroot((3x^4)+5) dx

All you need to do is set

Originally Posted by helmszee

antidifferentiation

sin^2xcos^2x dx

Now define for the remaining integral use the formula

Originally Posted by helmszee

also could some one check the antidifferentiation of

sin3xcos4x dx

From



and

Derive a formula for

Originally Posted by helmszee

antidifferentiation of

(x^3)/squareroot((3x^4)+5) dx

i know the formula 1/square root(1 - x^2) but that doesnt help me.

Observe .

Therefore:



RonL

Originally Posted by helmszee

can some one please check this one for me

antidifferentiation

sin^2xcos^2x dx

i got 1/3sinx^3 - 1/2sinx^4 + 1/5sin^5 + C

not very confident with this answer

Have you tried differentiating it to see if you get what you started with?

RonL

antidifferentiation of sin3xcos4x dx

using sinaxcosbx = 1/2sin(a-b)x + 1/2sin(a+b)x
=1/2sin(3-4)x + 1/2sin(3+4)x

-1/2sinx + 1/2sin7x dx

1/2cosx - 1/14cos7x + C

Which can't be simplified... i dont think..

as for antifdifferentiation of sin^2xcos^2x dx

let u = sinx
so du = cosx dx

antidif sin^2xcos^2x dx = sin^2xcos^2xcosx dx
= sin^2x(1-sinx)^2cosx dx
= u^2(1-u)2 du
= u^2(1-2u +u^2)
= (u^2 - 2u^3 + u^4) du
= 1/3u^3 - 1/2u^4 + 1/5u^5 + C
= 1/3sinx^3 - 1/2sinx^4 + 1/5sin^5 + C

hope that makes sence wen ever i used the math wrap text thing it always stuffs it up

Originally Posted by helmszee

antidifferentiation of sin3xcos4x dx

using sinaxcosbx = 1/2sin(a-b)x + 1/2sin(a+b)x
=1/2sin(3-4)x + 1/2sin(3+4)x

-1/2sinx + 1/2sin7x dx

1/2cosx - 1/14cos7x + C

Which can't be simplified... i dont think..

that is correct

as for antifdifferentiation of sin^2xcos^2x dx

let u = sinx
so du = cosx dx

antidif sin^2xcos^2x dx = sin^2xcos^2xcosx dx
...

here is your mistake, the two sides are not equal. you have cos^2(x) on the left but cos^3(x) on the right. follow the suggestion in post #2