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

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.

• Could you please try to chose a more expressive title.
– quid
Commented Jul 26, 2020 at 19:03
• I am sorry. This is my first time at stack exchange and also English is not my native language.
– user797330
Commented Jul 26, 2020 at 19:06
• For typesetting you can see math.meta.stackexchange.com/questions/5020/… For the title issue just try to say something a bit more detailed. Already saying it is about integrals of trigonometric functions would be better. Following those might avoid some pitfalls.
– quid
Commented Jul 26, 2020 at 19:22
• Since Sameer Baheti's answer shows this is true for integer $a$, you might try a small fraction for $a$ and see what happens. Commented Jul 26, 2020 at 19:49
• @Sameer titles that are only a formula are not ideal because it is difficult to "right click" them to open in new tab.
– quid
Commented Jul 26, 2020 at 21:08

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*}

• Yep, thank you a lot
– user797330
Commented Jul 26, 2020 at 19:46
• @YejatKrishna Check this link as per Marty cohen's answer. Your question is limited to integer $a$. Commented Jul 26, 2020 at 20:07

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.

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!

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.

• Thanks, I don't know why I don't use WA. I cross-checked your claim, and I guess I will use WA from now on instead of trying hard to prove something possibly incorrect. +1 btw. Commented Jul 26, 2020 at 20:05