## anonymous one year ago Use a substitution to find....

1. anonymous

|dw:1437224954877:dw|

2. anonymous

So i choose u=cos4t Derrivative gives du/dt=−4sin4t Then Rewrite sin4t dt=du/4 and finally i got to this 1/4∫sin^2 u du

3. UsukiDoll

I think it's one of those trigonometric .... where if the exponent is even and the exponent is odd...I forgot, but it does take long

4. UsukiDoll

ah trigonometric integrals

5. UsukiDoll

bu the problem is.. we have an exponent that's even (the sine_ and the exponent that's odd (cosine) and I can't find a good source at the moment that explains the rules for it nugh.

6. UsukiDoll

I think there are 3 cases one with all even exponents one with all odd exponents one with even and odd. and I hated these. I think I missed lecture eons ago

7. anonymous

The answer in the book is (Sin^3 * 4t)/ 4 + constant

8. UsukiDoll

@peachpi it's getting late on my end.. can't think x.x

9. anonymous

I was thinking the trig substitution as well but it looks like you can use straight u-sub. u = sin (4t) du = 4 cos (4t) dt ¼ du = cos (4t) dt $\frac{ 1 }{ 4 }\int\limits_{}u^2 du$

10. UsukiDoll

yeah that's what came to my mind first but then I saw the format of the question and I was like does that trig integral stuff come into play?

11. anonymous

whered u get sin(4t) from?

12. anonymous

that's in the original integral. sin² 4t = (sin 4t)²

13. UsukiDoll

|dw:1437225705494:dw|

14. anonymous

ohh i didnt know u could break it down like that. Alright let my try this again then.

15. UsukiDoll

yeah you can rewrite that sine like that I find it annoying when books have it in that notation so I change it

16. UsukiDoll

then I can see the u sub clearly...

17. anonymous

alright let me see if i can solve this.

18. anonymous

Yup I got the correct answer as the book now. I didnt understand u could change it like that. THanks guys!