# Thread: Integration by reduction: stuck hard

1. ## Integration by reduction: stuck hard

Hello people
I have been trying for hours to find the reduction formulae for this integrations with futility. Help would be hugely appreciated.
I am at a loss when trying to integrate these expressions by parts. While I decrease the degree of one part, the degree of another part increases so I am held impasse.
I need advise on integration by parts, especially on these types of sums.
Here are the problems:
(a) $\displaystyle \int^1_0 \frac{1}{(1+x^2)^n} dx$
(b) $\displaystyle \int^1_0 x^{n-2} \sqrt(1-x) dx$
The solutions:
(a) $\displaystyle 2(n-1)I_n=2^{1-n} + (2n-3) I_{n-1}$
(b) $\displaystyle I_n=\frac{n}{n+3} I_{n-2}$

2. $\displaystyle \int\limits_0^1 {\frac{1} {{\left( {1 + x^2 } \right)^n }}} dx = \left. {\frac{x} {{\left( {1 + x^2 } \right)^n }}} \right|_0^1 + 2n\int\limits_0^1 {\frac{{x^2 }} {{\left( {1 + x^2 } \right)^{n + 1} }}} dx =$

$\displaystyle = 2^{ - n} + 2n\int\limits_0^1 {\left( {1 - \frac{1} {{1 + x^2 }}} \right)\left( {\frac{1} {{\left( {1 + x^2 } \right)^n }}} \right)} dx =$

$\displaystyle \begin{gathered} = 2^{ - n} + 2n\left( {I_n - I_{n + 1} } \right) \hfill \\ \Rightarrow I_n = 2^{ - n} + 2n\left( {I_n - I_{n + 1} } \right) \hfill \\ \Leftrightarrow \left( {1 - 2n} \right)I_n = 2^{ - n} - 2nI_{n + 1} \hfill \\ \Leftrightarrow \left( {3 - 2n} \right)I_{n - 1} = 2^{1 - n} - 2\left( {n - 1} \right)I_n \hfill \\ \end{gathered}$

b.

$\displaystyle \int\limits_0^1 {x^{n - 2} \sqrt {1 - x} dx} = \left. { - \frac{2} {3}\left( {1 - x} \right)^{\frac{3} {2}} x^{n - 2} } \right|_0^1 + \frac{{2\left( {n - 2} \right)}} {3}\int\limits_0^1 {\left( {1 - x} \right)^{\frac{3} {2}} x^{n - 3} dx = }$

$\displaystyle \begin{gathered} = \frac{{2\left( {n - 2} \right)}} {3}\int\limits_0^1 {\left( {1 - x} \right)x^{n - 3} \sqrt {1 - x} dx = } \frac{{2\left( {n - 2} \right)}} {3}\left( {I_{n - 1} - I_n } \right) \hfill \\ \Leftrightarrow \left( {1 + \frac{{2\left( {n - 2} \right)}} {3}} \right)I_n = \frac{{2\left( {n - 2} \right)}} {3}I_{n - 1} \hfill \\ \end{gathered}$

...

3. Thanks a lot. These are really tricky, and i am just a novice.
I have a question on your working of (a), how do you convert I_{n+1} to I_{n-1} at the last steps:

I have spent another hour trying to understand this step and failing miserably. Dear me. Help me.

4. It's like when having $\displaystyle I_n=n\implies I_{n+1}=n+1,$ in this case, Peritus took $\displaystyle I_{n}$ and turned that as $\displaystyle I_{n-1}.$