math help

[CellCoree]

\int sin^3(4x)cos^{10}(4x) dx

ok i know that i need to borrow the a sin, cause the sin has an odd power


\int sin^2(4x)cos^{10}(4x) sin(4x)dx

ok used one of the trig ids. on sin^2(4x)

= \int (1-cos^2(4x)) * cos^{10}(4x) * sin(4x) dx

=\int cos^{10}(4x)*sin(4x)dx - \int cos^{12}(4x)cos^2(4x)sin(4x)dx

ok so i start to integrate the left side first...

\int sin(4x)cos^{10}(4x)dx
u=cos(4x)
du=-4sin(4x)dx
dx=-1/4 <-- needs to be balanced right? man im so confused on balancing, what am i suppose to balance anyway? i usally guess it lol, which is not good. please help me with this also

so...

\int -1/4 du * u^{10}
-1/4 \int u^{10} * du

now to find the anti-derv.

-1/4(u^{11}/11)

sub back in for u

-1/4(cos(4x)^{11}/11)

so this is what the problem looks right now...

-1/4(cos(4x)^{11}/11) - \int cos^{12}(4x)cos^2(4x)sin(4x)dx

ok how would i do the right side? it's so huge, so i use trig ids. agian? but am i doing it right so far?

[ehild]

\int sin^3(4x)cos^{10}(4x) dx

ok i know that i need to borrow the a sin, cause the sin has an odd power


\int sin^2(4x)cos^{10}(4x) sin(4x)dx

ok used one of the trig ids. on sin^2(4x)

= \int (1-cos^2(4x)) * cos^{10}(4x) * sin(4x) dx

=\int cos^{10}(4x)*sin(4x)dx - \int cos^{12}(4x)cos^2(4x)sin(4x)dx


The cos^2(4x) is superfluous in the second integral. It is just
cos^{12}(4x)sin(4x)dx
and you can do the same you did with the first one.

ehild

[Gza]

u=cos(4x)

du=-4sin(4x)dx

dx=-1/4 <-- needs to be balanced right? man im so confused on balancing, what am i suppose to balance anyway? i usally guess it lol, which is not good. please help me with this also

I'm not too sure what you mean by "balancing." What you are really doing is differentiatiating u=cos(4x) with respect to x, giving \frac{du}{dx}=-4sin(4x) , "multiplying through" by dx giving du=-4sin(4x)dx, and then simply solving the equation in terms of dx and then substituting into your original integral in \int sin(4x)cos^{10}(4x)dx .

ok how would i do the right side? it's so huge, so i use trig ids. agian? but am i doing it right so far?

You shouldn't have too much trouble with this integral since you were able to do the other one. And yes, your work looks fine to me. Post again if you need help with this integral as well.

[CellCoree]

The cos^2(4x) is superfluous in the second integral. It is just
cos^{12}(4x)sin(4x)dx
and you can do the same you did with the first one.

ehild


why did you remove cos^2(4x)?

i dont get it

for example:

= \int (1-cos^2(4x)) * cos^{10}(4x) * sin(4x) dx
you use 1 to multiply cos^{10}(4x) * sin(4x) dx then you use
cos^2(4x) to multiply cos^{10}(4x) * sin(4x) dx
right? that's how i got cos^{12}(4x)cos^2(4x)sin(4x)dx... am i doing something wrong?

[PureEnergy]

cos^{2}(4x)*cos^{10}(4x) * sin(4x) dx = cos^{12}(4x)sin(4x)dx

I believe the mistake you are making is multiplying cos^{2}(4x) once with the cos^{10}(4x) and another time with sin(4x) dx . This is incorrect because the two latter terms are multipled together.

[CellCoree]

thank you everyone, that really helped. im so glad i know how to do these problems now, it feels really goood. thank you agian