The option(s) with the values of a and L that satisfy the following equation is (are)
`(int_0^(4pi)e^t(sin^6at+cos^4at)dt)/(int_0^pi e^t(sin^6at+cos^4at)dt)=L`

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 `int_(0)^(pi)e^(t)(sin^(6)at+cos^(4)at)dt=I_(1)`
and `I_(2)=int_(pi)^(2pi)e^(t)(sin^(6)at+cos^(4)at)dt`
Put `t=x+pi` in `I_(2)`
`:. dt=dx`
For `a=2` and `a=4`
`:.int_(0)^(pi)e^(x+n)(sin^(6)ax+cos^(4)ax)dt=e^(pi)I_(1)`
Similarly,` int_(2pi)^(3pi)e^(t)(sin^(6)at+cos^(4)at)dt=e^(2pi)I_(1)`
and `int_(3pi)^(4pi)e^(t)(sin^(6)at+cos^(4)at)dt=e^(3pi)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_(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)/(int_(0)^(pi)e^(t)(sin^(6)at+cos^(4)at)dt)`
`=1+e^(pi)+e^(2pi)+e^(3pi)`
`=(e^(4pi)-1)/(e^(pi)-1)`