`I_10=int_0^(pi/2)x^(10)sinx dx` then `I_10+90I_8` is (A) `10(pi/2)^6` (B) `10(pi/2)^9` (C) `10(pi/2)^8` (D) `10(pi/2)^7`

Explanation

Given,

`I_(10)=int _0^(pi)/2 x^(10)sinxdx`


`I_(10)=[x^(10)int_0^(pi)/2 (sinx)]-int _0^(pi)/2[d/(dx) x^(10) int (sinx)dx]dx`


`I_(10)=[x^(10)(-cosx)]_0^(pi)/2-int _0^(pi)/2[10 x^9 int (sinx)dx]dx`


`I_(10)=[x^(10)(-cosx)]_0^(pi)/2-int _0^(pi)/2[10 x^9 -cosx]dx`
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