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Question

If I1=π0cosx(x+2)2 dx, I2=π/20cosxsinx(x+1) dx and λI2=2μ+π+kγI1, then

A
λ=4
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B
μ=2
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C
k=1
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D
γ=2
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Solution

The correct option is D γ=2
I1=π0cosx(x+2)2 dx
Put x=2t
I1=π/20cos2t2(t+1)2 dt
=[cos2t2(t+1)]π/20+π/202sin2t2(t+1) dt
I1=12+π+12π/204sintcost2(t+1) dt
2I1=22+π+14I2
4I2=22+π+12I1

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