Given information:

The integral is ∫0π2sinnx=n−1n∫0π2sinn−2xdx .

Proof: to proof, first use the reduction formula and then substitute,

Since, the reduction formula,

  ∫0π2sinnxdx=[−1nsinn−1xcosx]0π2+∫0π2(n−1)nsinn−2xdx∫0π2sinnxdx=0+∫0π2(n−1)nsinn−2xdx  [substitute and simplify]∫0π2sinnxdx=(n−1)n∫0π2sinn−2xdx

Given information:

The integral is ∫0π2sin3xdx and ∫0π2sin5xdx .

Calculation: to find the value of integral, first substitute in the formula and then simplify,

  ∫0π2sin3x=3−13∫0π2sin3−2xdx   [substitute n=3]∫0π2sin3x=23[−cosx]0π2            [integral]∫0π2sin3x=23(0+1)                   [substitute and simplify]∫0π2sin3x=23∫0π2sin5x=5−15∫0π2sin5−2xdx   [substitute n=3]∫0π2sin5x=45(23)        [simplify]∫0π2sin5x=815

Thus, the integral values are 23 and 815 .

Given information:

The integral is ∫0π2sin2n+1dx=2·4·6· · ·2n3·5·7· · ·(2n+1) .

Proof: to proof, use the part (a) and then simplify,

  ∫0π2sinnx=n−1n∫0π2sinn−2xdx∫0π2sin2n+1x=2n+1−12n+1∫0π2sin2n+1−2xdx  [substitute n=2n+1]∫0π2sin2n+1dx=2·4·6· · ·2n3·5·7· · ·(2n+1)    [by expanding and simplify]