The option(s) with the values of `aa n dL` that satisfy the following equation is (are) `(int0 4pie^t(s in^6a t+cos^4a t)dt)/(int0pie^t(s in^6a t+cos^4a t)dt)=L` `a=2,L=(e^(4pi)-1)/(e^(pi)-1)` (b) `a=2,L=(e^(4pi)+1)/(e^(pi)+1)` `a=4,L=(e^(4pi)-1)/(e^(pi)-1)` (d) `a=4,L=(e^(4pi)+1)/(e^(pi)+1)`

Options

A. `a=2, L=(e^(4pi)-1)/(e^(pi)-1)`
B. `a=2, L=(e^(4pi)+1)/(e^(pi)+1)`
C. `a=4, L=(e^(4pi)-1)/(e^(pi)-1)`
D. `a=4, L=(e^(4pi)+1)/(e^(pi)+1)`

Answer

A, C

Explanation

Let `I_(1)= int_(0)^(4pi)e^(t)(sin^(6)at +cos^(4) at) dt`
`=int_(0)^(pi)e^(t)(sin^(6) at +cos^(4) at )dt`
` +int_(pi)^(2pi)e^(t) (sin^(6) at +cos^(4) at )dt`
` +int_(2pi)^(3pi)e^(t) (sin^(6) at +cos^(4) at )dt`
` +int_(3pi)^(4pi)e^(t) (sin^(6) at +cos^(4) at )dt`
`:. I_(1)=I_(2)+I_(3)+I_(4)+I_(5)` . . . (i)
Now , `I_(3)=int_(pi)^(2pi)e^(t)(sin^(6) at + cos^(4) at)dt`
Put `t=pi+xrArr dt = dx`
`:. I_(3) = int_(0)^(pi)e^(pi+x)*(sin^(6) at + cos^(4) at ) dt = e ^(pi) * I_(2)` . . . (ii)
Now , `:. I_(4) = int_(2pi)^(3pi)e^(t)*(sin^(6) at + cos^(4) at ) dt `
Put `t= 2pi +xrArr dt = dx`
`:. I_(4)=int_(0)^(pi)e^(x+2pi)(sin^(6)at+cos^(4)at)dt = e^(2pi)*I_(2)` .. . (iii)
and `I_(5)= int_(3pi)^(4pi)e^(t)(sin^(6) at cos^(4)at)dt`
Put `t=3pi+x`
`:. I_(5)=e^(3pi+x)(sin^(6) at + cos^(4) at ) dt = e^(3pi)* I_(2)` . . . (iv)
Form Eqs . (i) , (ii) , (iii) and (iv) , we get
`I_(1)=I_(2)+e^(pi)*I_(2)+e^(3pi)*I_(2)=(1+e^(pi)+e^(2pi)+e^(3pi)) I_(2)`
` L= :.(int _(0)^(4pi)e^(t)(sin^(6)at + cos^(4) at)dt)/(int_(0)^(pi)e^(t)(sin^(6)at+cos^(4) at)dt)`
`=(1+e^(pi)+e^(2pi)+e^(3pi))`
`= (1*(e^(4pi)-1))/(e^(pi-1)) "for" a in R`