Use a substitution to find....

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Use a substitution to find....

Mathematics
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|dw:1437224954877:dw|
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
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

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ah trigonometric integrals
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.
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
The answer in the book is (Sin^3 * 4t)/ 4 + constant
@peachpi it's getting late on my end.. can't think x.x
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 \[\frac{ 1 }{ 4 }\int\limits_{}u^2 du\]
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?
whered u get sin(4t) from?
that's in the original integral. sin² 4t = (sin 4t)²
|dw:1437225705494:dw|
ohh i didnt know u could break it down like that. Alright let my try this again then.
yeah you can rewrite that sine like that I find it annoying when books have it in that notation so I change it
then I can see the u sub clearly...
alright let me see if i can solve this.
Yup I got the correct answer as the book now. I didnt understand u could change it like that. THanks guys!

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