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0-pi-0-2sin-1-rsin-r-dr-d-




Question Number 98744 by  M±th+et+s last updated on 15/Jun/20
∫_0 ^π ∫_0 ^(2sinθ) (1+rsinθ)r dr dθ
0π02sinθ(1+rsinθ)rdrdθ
Commented by bemath last updated on 16/Jun/20
= ∫_0 ^π [ (1/2)r^2 +(1/3)r^3 sin θ]_0 ^(2sin θ) dθ  = ∫_0 ^π (2sin^2 θ+(8/3)sin^4 θ) dθ  = ∫_0 ^π  (1−cos 2θ+(8/3)((1/2)−(1/2)cos 2θ)^2  dθ  = ∫_0 ^π (1−cos 2θ+(8/3)((1/4)−(1/2)cos 2θ+(1/4)((1/2)+(1/2)cos 4θ)) dθ  = ∫_0 ^π (1−cos 2θ+1 −(4/3)cos 2θ+(1/3)cos 4θ ) dθ  = [ 2θ−(7/6)sin 2θ +(1/(12))sin 4θ ]_0 ^π   = 2π ■
=π0[12r2+13r3sinθ]02sinθdθ=π0(2sin2θ+83sin4θ)dθ=π0(1cos2θ+83(1212cos2θ)2dθ=π0(1cos2θ+83(1412cos2θ+14(12+12cos4θ))dθ=π0(1cos2θ+143cos2θ+13cos4θ)dθ=[2θ76sin2θ+112sin4θ]0π=2π◼
Commented by  M±th+et+s last updated on 16/Jun/20
thank you sir
thankyousir

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