If I1=π∫0cosx(x+2)2dx,I2=π/2∫0cosxsinx(x+1)dx and λI2=2μ+π+k−γI1, 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 Dγ=2 I1=π∫0cosx(x+2)2dx
Put x=2t I1=π/2∫0cos2t2(t+1)2dt =[−cos2t2(t+1)]π/20+π/2∫0−2sin2t2(t+1)dt ⇒I1=12+π+12−π/2∫04sintcost2(t+1)dt ⇒2I1=22+π+1−4I2 ⇒4I2=22+π+1−2I1