need help with an indefinite integral
i need to practice on integrals so i'll help :P
can someone explain how\[\int\limits_{?}^{?}1/\theta^2 \cos(1/\theta)d \theta becomes -\sin(1/\theta)+c\]
I couldn't find the integral symbol without bounds so I just put them blank
lol you just have to erase the ? anyway.... let me try to fix this.... \(\large \int \frac{1}{\theta ^2} \cos (\frac {1}{\theta} ) d \theta = -\sin (\frac{1}{\theta} ) + C \) that's the question correct?
\[\int \frac{\cos(\frac{1}{\theta})}{\theta^2}d\theta\]
i will be quiet, just wanted to write it
the one that igbasallote posted
lol ok
lol
well then i'll give you a hint... let u = \(\large \frac{1}{\theta}\) du = \( \large \theta ^{-1} = -1(\theta ^{-2}) = - \frac{1}{\theta ^2}\) can you see how to do it now?
:O @Zarkon you're mod too??? wow congrats!
ty
@Zarkon congrats on being a mod. damn everyone but me? jealous
what is mod?
maybe in the next draft pick
haha lol yeah satellite i noticed too =)) you deserve to be one after all your contributions to this site ^_^
@chrisplusian mods are moderators kind of the ones who overlook the site..global moderators have great prestige because it means they are the best in their subject...anyway back to your question :D
you guys/ girls are awesome. I can't find help at my $240 a credit hour school so I come here:)
do you know how to go from that substitution @chrisplusian ?
yes i have it. because du is one over theta squared, d theta. and when I differentiate theta to the negative first it becomes one over d thaeta squared times d theta. I would have never seen that on my own. Thanks
@satellite73 maybe :)
it just takes practice :DDD just remember that when substituting..ALWAYS substitute the more complicated one
probably doesn't help that i am still trying to stay relatively anonymous
what makes it more complicated? the fact that it is the argument of a trig function?
yup :DD
by the way...i think the frontrunners for the next mod picking is satellite and hero..well i think they're the most deserving ones haha
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