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Question

The integral π/4π/6dxsin2x(tan5x+cot5x) equals :
  1. 110[π4tan1(193)]
  2. 15[π4tan1(133)]
  3. π40
  4. 120[tan1(193)]


Solution

The correct option is A 110[π4tan1(193)]
I=π/4π/6dxsin2x(tan5x+cot5x)
I=π/4π/6(1+tan2x)dx2tanx(tan5x+1tan5x)

Put tan x=t⎪ ⎪ ⎪⎪ ⎪ ⎪t=tanπ6=13t=tanπ4=1
sec2x dx=dt
(1+tan2x)dx=dt

I=11/312t(t5+1t5)dt
I=11/3t42(t10+1)dt

Put t5=u5t4dt=du
I=110135/21u2+1du

I=[110tan1u]135/2

=110[π4tan135/2]

=110[π4tan1(193)]

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