## anonymous 4 years ago sin^3x + cos^4x

1. anonymous

what is the problem here?

2. anonymous

how to integrate, sorry!

3. anonymous

Well, if the power of sine is odd and positive , what you want to do is lop of one of the sine factor, put it to the right of the rest of the expression and convert the remaining (even) sine factors to cosines withe the Pythagorean identity , and then integrate with the substitution method where u=cos(x)

4. anonymous

|dw:1325279411322:dw|

5. anonymous

oh, ok, then, thanks.

6. anonymous

hold on let me finish it for you: $\int sin^3(x)cos^4(x)dx =\int sin^2(x)cos^4(x)sin(x)dx$

7. anonymous

:D

8. anonymous

$\int sin^2(x)cos^4(x)sin(x)dx=\int (1-cos^2(x))cos^4(x)sin(x)dx$

9. anonymous

$\int (1-cos^2(x))cos^4(x)sin(x)dx=\int (cos^4(x)-cos^6(x))sin(x)dx$

10. anonymous

$u=cos(x)\rightarrow \frac{du}{dx}=-sin(x)\rightarrow du=-sin(x)dx\rightarrow \frac{-du}{sin(x)}=dx$

11. anonymous

$\int (cos^4(x)-cos^6(x))sin(x)dx=-\int (u^4-u^6)du$

12. anonymous

$=-\int (u^4-u^6)du=-\frac{1}{5}u^5+\frac{1}{7}u^7+C$

13. anonymous

$=\frac{1}{7}cos^7(x)-\frac{1}{5}cos^5(x)+C$

14. anonymous

15. anonymous

so simple. thanks!

16. anonymous

well, its simple once you see it