1. ## 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

2. Originally Posted by helmszee
antidifferentiation of

(x^3)/squareroot((3x^4)+5) dx
$\int\frac{x^3}{\sqrt{3x^4+5}}\,dx$

All you need to do is set $u=\sqrt{3x^4+5}$

Originally Posted by helmszee
antidifferentiation

sin^2xcos^2x dx
$\int\sin^2x\cos^2x\,dx=\frac14\int\sin^22x$

Now define $u=2x,$ for the remaining integral use the formula $\cos2v=\cos^2v-\sin^2v.$

Originally Posted by helmszee
also could some one check the antidifferentiation of

sin3xcos4x dx
From

$\sin (\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

and $\sin (\alpha - \beta ) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$

Derive a formula for $\sin \alpha \cos \beta$

3. 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 $\frac{d}{dx}~\sqrt{3x^4+5}=\frac{(1/2)(12x^3)}{\sqrt{3x^4+5}}$.

Therefore:

$
\int \frac{x^3}{\sqrt{3 x^4+5}} ~dx = \frac{1}{6}~\sqrt{3x^4+5}+C
$

RonL

4. 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

5. 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

6. 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