Integral equation

[kidmo87]

Hello everyone, I'm a little lost at a certain part of this integral equation. The part that I boxed in red is what im having problems with, mainly the first like of the red box. I was hoping someone could explain to me how to do it. Is there a formula being used. I was absent in class because I was sick, thanks.

 

[HallsofIvy]

Hello everyone, I'm a little lost at a certain part of this integral equation. The part that I boxed in red is what im having problems with, mainly the first like of the red box. I was hoping someone could explain to me how to do it. Is there a formula being used. I was absent in class because I was sick, thanks.View attachment 2719

First, that is not an "integral equation", just an integral you are asked to evaluate.

Second, you have \(\displaystyle \int (1- 2sin(x)+ sin^4(x)) cos(x) dx\) (you are missing the last "(x)".)

To get the first line in the red box, do two things:
1) Use the "distributive law": \(\displaystyle (1- 2sin(x)+ sin^4(x)) cos(x)= cos(x)- 2sin(x)cos(x)+ sin^4(x)cos(x)\).
2) Use the fact that \(\displaystyle \int A+ B dx= \int A dx+ \int B dx\):
\(\displaystyle \int (1- 2sin(x)+ sin^4(x)) cos(x) dx= \int (cos(x)- 2sin(x)cos(x)+ sin^4(x)cos(x))dx= \int cos(x)dx- 2\int sin(x)cos(x)dx+ \int sin^4(x)cos(x)dx\)
In each of those, use the substitution u= sin(x) so that du= cos(x)dx and that becomes
\(\displaystyle \int du- 2\int u du+ \int u^4 du\)

 

[kidmo87]

Thank you soo much!!