#Integration way harder then what it looks like
35 messages · Page 1 of 1 (latest)
Can* match it
Did you try partial fraction decomposition?
i think it is the only way
@limpid sentinel this is the way
you will have to solve this equation
by the way, which book are you using ?
another way maybe a bit easier is to substitute x=2tan(t)
You get cos⁶t. Then express cos(t)^3 in terms of cos(3t) and square it. So, you easily get
32cos⁶t=cos(6t)+6cos(4t)+15cos(2t)+10. Integrate it. Then express it back in cos(2t), sin(2t) which then is easy in terms of tan(t)=x/2. So, yes, this should be way shorter on paper.
you can also do it by induction with $I_p = \int \frac{dx}{(4+x^2)^p}$
bloubbloub
also you can't decompose it further with PFD
i did not thinked about it but good idea!
yes or use integration by parts, there is a formula $I_n=\int_{}^{}cos^{n}(x)dx=\frac{cos^{n-1}(x)sin(x)-(n-1)\cdot I_{n-2}}{n}$
barann0056
my bad it is :
$In=\int{}^{}cos^{n}(x)dx=\frac{cos^{n-1}(x)sin(x)+(n-1)\cdot I_{n-2}}{n}$
barann0056
I did try that cause I saw x^2+4 but it’s didn’t work it was just too lengthly and I couldn’t find a way
It’s given by our teacher in homework sheet I don’t know his source
Sir himself was unable to solve it in doubt class
He tried tan2x substitution and by part too but it was too lengthily and was going nowhere
i mean it is pretty hard if you did not worked a lot on computation
so its okay if you don't succed this one
good luck for your calculus class
Why, it should be no more than a page of formulas.
It perfectly works.
because you are not lazy
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@limpid sentinel type .close
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Post marked as solved by @limpid sentinel.
Use .unsolved if this was a mistake.
Yea I got it already now
Same thing I did
Still thanks
Still very lengthy took very long