The values of constants in the Equation. $$ \frac{\int_0^{4\pi} e^t(\sin^6at+\cos^4at)\,dt}{\int_0^\pi e^t(\sin^6at+\cos^4at)\,dt}= L, $$
the question asks the value of $a$ and $L$.
My friend solved it by differentiating, but i didn't understood a single thing. May be it is a very simple one, but i don't know how to solve it. 
 A: Notice that the numerator can be written as
$$\int_{0}^{4\pi}e^t(\sin^{6}at+\cos^{4}at)dt=\int_{0}^{\pi}e^t(\sin^{6}at+\cos^{4}at)dt+\int_{\pi}^{2\pi}e^t(\sin^{6}at+\cos^{4}at)dt+\int_{2\pi}^{3\pi}e^t(\sin^{6}at+\cos^{4}at)dt+\int_{3\pi}^{4\pi}e^t(\sin^{6}at+\cos^{4}at)dt$$
Now substitute $t=k+\pi$ in the second integral on the R.H.S, to get $$\int_{\pi}^{2\pi}e^t(\sin^{6}at+\cos^{4}at)dt=\int_{0}^{\pi}e^{k+\pi}(\sin^{6}ak+\cos^{4}ak)dk$$
because $\sin^6ak=\sin^6(ak+\pi)$ and same for $\cos^4ak$
$$=e^\pi \times \int_{0}^{\pi}e^k(\sin^{6}ak+\cos^{4}ak)dk$$
Similarly for the other two integrals substitute $t=k+2\pi$ and $t=k+3\pi$ to get 
$$\frac{\int_{0}^{4\pi}e^t(\sin^{6}at+\cos^{4}at)dt}{\int_{0}^{\pi}e^t(\sin^{6}at+\cos^{4}at)dt}=1+e^\pi+e^{2\pi}+e^{3\pi}$$
$$=\color{red} {\frac{e^{4\pi}-1}{e^\pi-1}}$$
Why the particular substitutions?

These are done in order to make the limits of the integrals present in the  numerators and denominators the same while also keeping in mind the periodicity of the trigonometric functions.

A: HINT:

Use, one of the properties of the intergal:
$$\int\left[f(x)+g(x)\right]\space\text{d}x=\int f(x)\space\text{d}x+\int g(x)\space\text{d}x$$

So, we get that:
$$\int e^t\left(\sin^6(at)+\cos^4(at)\right)\space\text{d}t=\int e^t\sin^6(at)\space\text{d}t+\int e^t\cos^4(at)\space\text{d}t$$
Now, for $\int e^t\sin^6(at)\space\text{d}t$ use the reduction formula:
$$\int e^{at}\sin^n(bt)\space\text{d}t=\frac{e^{at}\sin^{n-1}(bt)(a\sin(bt)-bn\cos(bt))}{a^2+b^2n^2}+\frac{b^2(n-1)n\int e^{at}\sin^{n-2}bt)\space\text{d}t}{a^2+b^2n^2}$$
Use the trigonometric identity:
$$\sin^2(at)=\frac{1-\cos(2at)}{2}$$
Now, for $\int e^t\cos^4(at)\space\text{d}t$ use the reduction formula:
$$\int e^{at}\cos^n(bt)\space\text{d}t=\frac{bn\sin(bt)+a\cos(bt)}{a^2+b^2n^2}+\frac{b^2(n-1)n\int e^{at}\cos^{n-2}(bt)\space\text{d}t}{a^2+b^2n^2}$$
Use the trigonometric identity:
$$\cos^2(at)=\frac{1+\cos(2at)}{2}$$
A: We have$$\frac{\int_0^{4\pi} e^t(\sin^6at+\cos^4at)\,dt}{\int_0^\pi e^t(\sin^6at+\cos^4at)\,dt}= F(a)$$ It follows the equation $$F(a)=L$$
The calculation made by Nikunj makes us see that $F$ is independent of $a$.
Thus the answer is $a$  arbitrary and $L={\frac{e^{4\pi}-1}{e^\pi-1}}\approx 12.9512$
