- #1

- 51

- 7

**Summary:**Ιntegral calculation : (sin(x))^4 * (cos(x))^6

Hi all,

I tried to solve it, but I got stuck. An advice from my professor is to set: x=arctan(t)

Τhanks.

- Thread starter Michael_0039
- Start date

- #1

- 51

- 7

Hi all,

I tried to solve it, but I got stuck. An advice from my professor is to set: x=arctan(t)

Τhanks.

- #2

- 15,621

- 7,752

They are on my list of standard integrals.

- #3

fresh_42

Mentor

- 14,137

- 11,436

I have found the formula on Wikipedia.

- #4

- 51

- 7

Now it's in simple format:

Do you know a better - faster way to solve it ?

Thanks.

- #5

fresh_42

Mentor

- 14,137

- 11,436

$$

\int \sin^n x \cos^m x \,dx = -\dfrac{\sin^{n-1}x \cos^{m+1}x}{n+m} +\dfrac{n-1}{n+m}\int \sin^{n-2} x \cos^m x \,dx

$$

- #6

- 15,621

- 7,752

I would have started with ##\sin^4 x = (1 - \cos^2 x)^2 = 1 - 2\cos^2x + cos^4x##.Do you know a better - faster way to solve it ?

Thanks.

Then used parts to calculate the integral of the various powers of ##\cos x##.

This is so useful and common that I keep the recursion formula in my list of standard integrals.

- #7

- 15,621

- 7,752

A thing of beauty!

$$

\int \sin^n x \cos^m x \,dx = -\dfrac{\sin^{n-1}x \cos^{m+1}x}{n+m} +\dfrac{n-1}{n+m}\int \sin^{n-2} x \cos^m x \,dx

$$

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