# Thread: pls help to find integral

1. ## pls help to find integral

I have two questions. My idea was to use u substitution or integration by parts

i would appreciate the help.

2. ## Re: pls help to find integral

here has a form identical to your second integral.

It's recursive though. You'll need to write a small program to evaluate it.

For your first integral there's nothing fundamentally tricky about it. You can use binomial expansion to deal with the second term and then just multiply by the first. Each of the summands is easily integrated. Again a short program will take care of keeping track of all the terms.

There may be some super clever trick here I don't know about.

3. ## Re: pls help to find integral

Originally Posted by start5
I have two questions. My idea was to use u substitution or integration by parts

i would appreciate the help.
$\displaystyle\int {{{\sin }^{76}}(x){{\cos }^3}(x)dx = \int {{{\sin }^{76}}(x)\cos (x)dx} } - \int {{{\sin }^{78}}(x)\cos (x)dx}$

4. ## Re: pls help to find integral

For the first one, substitute $\displaystyle u = 1 + x^2 + x^4$. Then $\displaystyle du = (2x + 4x^3)\,dx$, which is just half of the first polynomial $\displaystyle 4x + 8x^3$.

- Hollywood

5. ## Re: pls help to find integral

oh well the trig integral becomes simple then as well.

let

$$u=sin(x)$$
$$du=cos(x)dx$$

6. ## Re: pls help to find integral

Originally Posted by romsek
oh well the trig integral becomes simple then as well.
let $$u=sin(x)$$ $$du=cos(x)dx$$
$$cos^2(x)=1-sin^2(x)$$
$$\int \left(u^{76}-u^{78}\right)du=\frac{u^{77}}{77}-\frac{u^{79}}{79}=$$
$$\frac{\sin^{77}(x)}{77}-\frac{\sin^{79}(x)}{79}$$
Why are we now giving out giving out complete solutions?

7. ## Re: pls help to find integral

fixed

of course the solution is in your quote and there's nothing I can do about that.