# evaluation line integral

i was reading special tutorial about evaluation line integral and author done following, he had to evaluate $$\int_{-1}^1 \cos(t)\,\sin^4(t)dt,$$ he did it by integration by part or denoted $u=\sin(t)$, $dv=\cos(t)dt$, got
$$\int_{-1}^1 u^4dv,$$ everything is clear still here but after integration he wrote it as
$$\frac{u^5}{5},$$ i know that $\int x^adx$ is equal $\frac{x^{a+1}}{a+1}$ but here $u^4 dv$ could not be just $\frac{u^5}{5}$, am i correct? i think it must be $$u^4v-v\int u^4 dt,$$ please tell me if i am wrong

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I have edited your post so that the mathematics is in LaTeX, please feel free to edit the post if I have accidentally changed your intended meaning. –  Zev Chonoles Jul 22 '11 at 7:21

It seems that what has happened here is just that $u$ and $v$ look awfully similar, and you have mistakenly thought the computation was an integration by parts (where we pick factors of the integrand, and label them $u$ (yoo) and $dv$ (dee vee)), when in fact it was a $u$-substitution (where we call some piece of the integrand $u$ (yoo), and express $du$ (dee yoo) in terms of our previous variable $t$ and its differential $dt$).

From user3196's explanation of the source below, here is a screenshot:

Indeed, when setting up the substitution, the "$u$" part was distinctively serifed, while the "$du$" part was not. This inconsistency in the handwriting of the creator of this video is definitely the cause for the confusion.

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♦ yes maybe i have made some mistake just here is link from youtube youtube.com/watch?v=fjEvsinvtnw –  dato datuashvili Jul 22 '11 at 7:36
♦ yes maybe i have made some mistake just here is link from youtube –  dato datuashvili Jul 22 '11 at 7:36

Looks to me like a typo. This is integration by substitution, not parts, and it should be $du=\cos t\,dt$.

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