paglia 4 years ago sin^3x + cos^4x

1. LagrangeSon678

what is the problem here?

2. paglia

how to integrate, sorry!

3. LagrangeSon678

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. LagrangeSon678

|dw:1325279411322:dw|

5. paglia

oh, ok, then, thanks.

6. LagrangeSon678

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. paglia

:D

8. LagrangeSon678

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

9. LagrangeSon678

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

10. LagrangeSon678

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

11. LagrangeSon678

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

12. LagrangeSon678

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

13. LagrangeSon678

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

14. LagrangeSon678

15. paglia

so simple. thanks!

16. LagrangeSon678

well, its simple once you see it