## paglia Group Title sin^3x + cos^4x 2 years ago 2 years ago

1. LagrangeSon678 Group Title

what is the problem here?

2. paglia Group Title

how to integrate, sorry!

3. LagrangeSon678 Group Title

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 Group Title

|dw:1325279411322:dw|

5. paglia Group Title

oh, ok, then, thanks.

6. LagrangeSon678 Group Title

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 Group Title

:D

8. LagrangeSon678 Group Title

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

9. LagrangeSon678 Group Title

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

10. LagrangeSon678 Group Title

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

11. LagrangeSon678 Group Title

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

12. LagrangeSon678 Group Title

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

13. LagrangeSon678 Group Title

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

14. LagrangeSon678 Group Title