If
$\large \rm I_n = \int_0^1 x(1-x^3)^n dx$

Prove that
$$\large \rm (3n+2)I_n = 3nI_{n-1}$$

Question

If 
$\large \rm I_n = \int_0^1 x(1-x^3)^n dx$

Prove that 
$$\large \rm (3n+2)I_n = 3nI_{n-1}$$

faiqraees · Answer

@ganeshie8

faiqraees · Answer

If someone can post the solution or the starting integral form, it will be fine. I will work the rest out myself.

danni_joanaveil2 · Answer

i hope this helps https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=9&cad=rja&uact=8&ved=0ahUKEwjdvfr7y9zOAhUBlh4KHWjaBVUQFgg7MAg&url=http%3A%2F%2Fdocslide.us%2Ftechnology%2Finstructors-solutions-manual-marion-thornton-classical-dynamics-of-particles-and-systems-5-th-ed.html&usg=AFQjCNHIZ0Qz3lr4EVJixAiT9HAbFBTUVQ

faiqraees · Answer

Nope it didn't help. It is calculus. The reference you gave was for mechanics.

danni_joanaveil2 · Answer

im sorry hun i dont know then :(

faiqraees · Answer

No problem

irishboy123 · Answer

$I_n = \int\limits_0^1 x (1-x^3)^n dx$

$I_n = \int\limits_0^1 x (1-x^3) (1-x^3)^{n-1} dx$

$I_n = \int\limits_0^1 x (1-x^3)^{n-1}  - x^4 (1-x^3)^{n-1} dx$

$I_n =  I_{n-1}  - \int\limits_0^1 x^4 (1-x^3)^{n-1} dx$

$I_n =  I_{n-1}  - \int\limits_0^1 x^2 .( x^2 (1-x^3)^{n-1}) dx$

IBP it now as per the hint 

i haven't actually done it but i think that should work

faiqraees · Answer

@Irishboy123 thank you very much

kainui · Answer

Fun looking problem

faiqraees · Answer

$\color{#0cbb34}{\text{Originally Posted by}}$ @Kainui
Fun looking problem
$\color{#0cbb34}{\text{End of Quote}}$
Only mathematicians can consider a problem fun.