RohitShubesh21
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recursive integration (1 Viewer)
- Thread starter RohitShubesh21
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5uckerberg
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Let's do this properly.
^{n}}dx=\int_{0}^{1}x^{2}\left(1+x^{2}\right)^{-n}dx)
After a painstaking battle, I have found that this question can easily get overcomplicated by the student.
The first step integrate
By doing so we will have^{-n}\right]_{0}^{1})
Sub in x=1 and we will have
and x=0 will give us 0.
Fear not have faith in yourself.
Now we will be differentiating^{-n})
The chain rule is your friend^{-n}=-2nx\left(1+x^{2}\right)^{-n-1}=-\frac{2nx}{\left(1+x^{2}\right)^{n+1}})
Multiply everything by
and we will have ^{n+1}})
. In conclusion we are workin g with
^{n}}dx=\int_{0}^{1}x^{2}\left(1+x^{2}\right)^{-n}dx=\frac{2^{-n}}{3}+\int_{0}^{1}\frac{2nx^{4}}{3\left(1+x^{2}\right)^{n+1}}dx)
Simply put, we have^{n+1}}dx)
.
Now, multiply by 3
^{n+1}}dx)
^{n+1}}dx)
Now what?
Keep an eye on this term^{n+1}}dx)
What can you do to get something like^{n}})
?
If you can solve that step you will obtain a huge morale boost.
^{n+1}}dx=\frac{2nx^{2}}{\left(1+x^{2}\right)^{n}}\times{\left(\frac{x^{2}}{1+x^{2}}\right))
^{n}}\times{\left(\frac{x^{2}}{1+x^{2}}\right)=\frac{2nx^{2}}{\left(1+x^{2}\right)^{n}}\times\left(1-\frac{1}{1+x^{2}}\right))
Let's finish this the final stretch.
^{n}}\times\left(1-\frac{1}{1+x^{2}}\right)=\frac{2nx^{2}}{\left(1+x^{2}\right)^{n}}-\frac{2nx^{2}}{\left(1+x^{2}\right)^{n+1}}=2nI_{n}-2nI_{n+1})
Combining it all together
.
Rearranging we will then have
I_{n})
After a painstaking battle, I have found that this question can easily get overcomplicated by the student.
The first step integrate
By doing so we will have
Sub in x=1 and we will have
Fear not have faith in yourself.
Now we will be differentiating
The chain rule is your friend
Multiply everything by
Simply put, we have
Now, multiply by 3
Now what?
Keep an eye on this term
What can you do to get something like
If you can solve that step you will obtain a huge morale boost.
Let's finish this the final stretch.
Combining it all together
Rearranging we will then have