# Math sin integral 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 I am 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?

Homework Helper
CellCoree said:
$$\int sin^3(4x)cos^{10}(4x) dx$$

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 I am 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
ehild said:
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 don't 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. I am so glad i know how to do these problems now, it feels really goood. thank you agian