Is the following true $\frac{\int_0^{4\pi}e^x(\sin^6 ax + \cos^4 ax)\,dx}{\int_0^{\pi}e^x(\sin^6 ax + \cos^4 ax)\,dx}=\frac{e^{4\pi}-1}{e^{\pi}-1}$?

Question

If $f(x)=e^x(\sin^6 ax + \cos^4 ax)$ where $a\in\mathbb Z$. Let $S_1$ be the area of the region bounded by $y = f (x)$, with x-axis and between the ordinates $x=0$, $x=4\pi$ , let $S_2$ be the area of the region bounded by $y =f(x)$, with x-axis and between the ordinates $x=0$ and $x=\pi$. Further let $S_1/S_2=A$

Now one possible conclusion has been provided to me as $A=(e^{4\pi}-1)/(e^{\pi}-1)$

Please provide me with suitable steps to arrive at this conclusion.

Answer 1

Prove that $$\frac{\int_0^{4\pi}e^x(\sin^6 ax + \cos^4 ax)\,dx}{\int_0^{\pi}e^x(\sin^6 ax + \cos^4 ax)\,dx}=\frac{e^{4\pi}-1}{e^{\pi}-1}$$, where $a\in \mathbb Z$.

\begin{align*} &\Rightarrow\int_0^{4\pi}e^x(\sin^6 ax + \cos^4 ax)\,dx\\ &=\int_0^{\pi}e^x(\sin^6 ax + \cos^4 ax)\,dx+\int_{\pi}^{2\pi}e^x(\sin^6 ax + \cos^4 ax)\,dx+\int_{2\pi}^{3\pi}e^x(\sin^6 ax + \cos^4 ax)\,dx+\int_{3\pi}^{4\pi}e^x(\sin^6 ax + \cos^4 ax)\,dx\\ &=\int_0^{\pi}e^x(\sin^6 ax + \cos^4 ax)\,dx+\int_{0}^{\pi}e^{x+\pi}(\sin^6 ax + \cos^4 ax)\,dx+\int_{0}^{\pi}e^{x+2\pi}(\sin^6 ax + \cos^4 ax)\,dx+\int_{0}^{\pi}e^{x+3\pi}(\sin^6 ax + \cos^4 ax)\,dx\\ &=\int_0^{\pi}(e^x+e^{x+\pi}+e^{x+2\pi}+e^{x+3\pi})(\sin^6 ax + \cos^4 ax)\,dx\\ &=(1+e^{\pi}+e^{2\pi}+e^{3\pi})\int_0^{\pi}e^x(\sin^6 ax + \cos^4 ax)\,dx\\ &=\frac{e^{4\pi}-1}{e^{\pi}-1}\int_0^{\pi}e^x(\sin^6 ax + \cos^4 ax)\,dx\\ \end{align*}

Answer 2

As @Integrand answered, using the power reduction formulae and integration by parts, we have $$\int e^x(\sin^6 (ax) + \cos^4 (ax))\,dx=$$ $$\frac{e^x }{32} \left(\frac{2 a \sin (2 a x)}{4 a^2+1}+\frac{40 a \sin (4 a x)}{16 a^2+1}-\frac{6 a \sin (6 a x)}{36 a^2+1}+\frac{\cos (2 a x)}{4 a^2+1}+\frac{10 \cos (4 a x)}{16 a^2+1}-\frac{\cos (6 a x)}{36 a^2+1}+22\right)$$

If $a$ is an integer, then $$I_n=\frac{\int_0^{n\pi}e^x(\sin^6 ax + \cos^4 ax)\,dx}{\int_0^{\pi}e^x(\sin^6 ax + \cos^4 ax)\,dx}=\sum_{k=0}^{n-1}e^{k \pi}=\frac{e^{n\pi }-1}{e^{\pi }-1}$$ as @Sameer Baheti already answered.

In any other case, this does not hold, as @marty cohen showed.

Answer 3

Hints:

• Use the Chebyshev formulas, also known as the power-reduction formulas, to turn powers into sums. For instance, $$ \cos^6(\theta)=\frac{1}{32} \left (-15 \cos(2 \theta) + 6 \cos(4 \theta) - \cos(6 \theta) + 10\right) $$
• Use integration by parts twice on integrals of the form $\int e^x \cos(m x)\,dx$ with $u=e^x$. You'll get a self-similar integral, which you can then isolate.

Hope that helps!

Answer 4

For $a=\frac12$, Wolfy says $\int_0^{4 π} e^x (sin^6(ax) + cos^4(ax)) dx = (e^{4 π} - 1)(61/80) $ and $\int_0^{ π} e^x (sin^6(ax) + cos^4(ax)) dx = (59e^{ π} - 61)/80 $, so it is not true for this.