## KarlaKalurky Group Title integrate: one year ago one year ago

1. KarlaKalurky Group Title

|dw:1352086212618:dw|

2. KarlaKalurky Group Title

let u=tan theta du= sec^2 d(theta)

3. KarlaKalurky Group Title

could it be: |dw:1352086779624:dw| ?

4. KingGeorge Group Title

Are you told to do a u-sub of $$u=\tan(\theta)$$?

5. KarlaKalurky Group Title

yes.,

6. KingGeorge Group Title

Well then, you get $\int \tan(\theta)^2\sec(\theta)^4 d\theta$With a u-sub of $$u=\tan(\theta)$$, we have $$du=\sec(\theta)^2d\theta$$, so it seems like we get$\int u^2\sec(\theta)^2 du$or$\int u^2 \frac{du^2}{d\theta}$

7. KingGeorge Group Title

However, if we start back at $$du=\sec(\theta)^2d\theta$$, and take the derivative again, we get $du^2=2\tan(\theta)\sec(\theta)^2d\theta^2.$Honestly, I'm not sure where we're supposed to go with that u-sub.

8. KarlaKalurky Group Title

|dw:1352087930624:dw| ?

9. KingGeorge Group Title

What I seem to be getting, is this. I'm not quite sure what to do with it.$\int u^2\frac{(du)^2}{d\theta}$

10. KarlaKalurky Group Title

uhm...just try to write your solution. ill appreciate it :)

11. KingGeorge Group Title

Hold on a second. I think I know what you need to do. Let $$u=\tan(\theta)$$. Then use the identity $$\sec(\theta)^2=\tan(\theta)^2+1$$. That means we now have$\large \int \tan(\theta)^2(\tan(\theta)^2+1)\sec(\theta)^2 d\theta$Now you make the u-sub.

12. KingGeorge Group Title

We get $\large \int u^2(u^2+1) \;du=\int u^4+u^2 \;du.$ This you should be able to integrate. Then just substitute $$u=\tan(\theta)$$ back in, and you're good.

13. KarlaKalurky Group Title

thanks a lot!! :)

14. KarlaKalurky Group Title

|dw:1352125444501:dw| |dw:1352125609551:dw|