The option (s) with the values of a and L that satisfy the following equation is (are) ∫04πet(sin6at+cos4at)dt∫0πet(sin6at+cos4at)dt=L?

We haveIn=∫nπn+1πetsin6at+cos4atdtsubstitute t=nπ+θ⇒In=enπ∫0πeθsin6aθ+cos4aθdθ∴∫04πetsin6at+cos4atdt=∫0π+∫π2π+∫2π3π+∫3π4π=1+eπ+e2π+e3π∫0πetsin6at+cos4atdt=e4π−1eπ−1∫0πetsin6at+cos4atdt∴∫04πetsin6at+cos4atdt∫0πetsin6at+cos4atdt=e4π−1eπ−1For any value of ‘a’ the above result holds.Options (1),(3) are correct.