Question Number 114689 by arkanmath7@gmail.com last updated on 20/Sep/20

∫xsin^n xdx

Answered by 1549442205PVT last updated on 20/Sep/20

Just give out reduction formula:  Integrating by parts we get:  I_n =∫xsin^n xdx=∫xsin^(n−1) x.sinxdx  =−xsin^(n−1) (x)cosx+∫cosx{sin^(n−1) x+x(n−1)sin^(n−2) xcosx}dx  =−xcosxsin^(n−1) x+∫sin^(n−1) xcosxdx  +(n−1)∫xsin^(n−2) x(1−sin^2 x)dx  =−xcosxsin^(n−1) x+((sin^n x)/n)+(n−1)(I_(n−2) −I_n )  ⇒(1+(n−1)I_n =−xcosxsin^(n−1) x+((sin^n x)/n)  +(n−1)I_(n−2)   ⇒I_n =((−xcosxsin^(n−1) x)/n)+((sin^n x)/n^2 )+((n−1)/n)I_(n−2)

Commented byarkanmath7@gmail.com last updated on 20/Sep/20

I solved it in same way but I stopped at  I_(n−2) . Is it true to let it as it is?

Commented by1549442205PVT last updated on 20/Sep/20

Just need above formula we will find  out I_n for ∀n∈N^∗   I_n ⇐I_(n−2) ⇐I_(n−4) ⇐....⇐I_1 (if n odd)  I_n ⇐I_(n−2) ⇐....⇐I_2 (if n even)  I_1 ,I_2 are simple cases