MHB Can You Evaluate This Definite Integral with Trig Functions?

AI Thread Summary
The discussion centers on evaluating the definite integral $\int_{-\pi}^{\pi} \frac{\sin nx}{(1+2^x)\sin x}\,dx$, with participants expressing confusion about the integrand's transformation and the introduction of the factor $2^x$. Questions arise regarding how this factor relates to the overall evaluation and the adjustments made to the bounds of integration. One participant indicates that they are awaiting further clarification from another member who has promised a proof. The conversation highlights the complexities involved in understanding the integral's manipulation and the need for additional explanations. Overall, the thread reflects a collaborative effort to clarify the evaluation process of the integral.
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Evaluate the definite integral $\displaystyle \int_{-\pi}^{\pi} \dfrac{\sin nx}{(1+2^x)\sin x}\,dx$, where $n$ is a natural number.
 
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$$I=\displaystyle \int_{-\pi}^{\pi} \dfrac{\sin nx}{(1+2^x)\sin x}\,dx $$

Use the substitution $t = -x$

$$I = \displaystyle \int_{-\pi}^{\pi} \dfrac{2^t\sin nt}{(1+2^t)\sin t}\,dt = \int_{-\pi}^{\pi} \dfrac{2^x\sin nx}{(1+2^x)\sin x}\,dx $$

Since $t$ is just a dummy variable and by summing the two formulas

$$2I =\int_{-\pi}^{\pi} \dfrac{\sin nx}{\sin x}\,dx = \int_{0}^{2\pi} \dfrac{\sin nx}{\sin x}\,dx $$

Let us consider two different cases when $n$ is even

$$I_k = \int^{2\pi}_0\frac{\sin(2kx)}{\sin(x)}\,dx$$

$$I_k-I_{k-1} = \int^{2\pi}_0\frac{\sin(2kx)-\sin(2kx-2x)}{\sin(x)}\,dx$$

$$I_k-I_{k-1} =2 \int^{2\pi}_0\frac{\sin(x)\cos(2kx-x)}{\sin(x)}\,dx = \int^{2\pi}_0\cos(2kx-x)\,dx = 0$$

Hence

$$I_k=I_{k-1} = I_1 = 2\int^{2\pi}_0 \cos(x) \,dx= 0 $$

If $n$ is odd , let

$$I_j = \int^{2\pi}_0\frac{\sin((2j+1)x)}{\sin(x)}\,dx$$

$$I_j-I_{j-1} = \int^{2\pi}_0\frac{\sin((2j+1)x)-\sin((2j-1)x)}{\sin(x)}\,dx$$

$$I_j-I_{j-1} = \int^{2\pi}_0\frac{\sin(x)\cos(2jx)}{\sin(x)}\,dx = 0$$

Similarily

$$I_j = I_{j-1} = I_0 = \int^{2\pi}_0 \frac{\sin(x)}{\sin(x)} \,dx= 2\pi$$

Finally we get that

$$I= \int_{-\pi}^{\pi} \dfrac{\sin nx}{(1+2^x)\sin x}\,dx= \begin{cases}
0 & \text{n is even} \\
\pi & \text{n is odd}
\end{cases}$$
 
Last edited:
ZaidAlyafey said:
$$I=\displaystyle \int_{-\pi}^{\pi} \dfrac{\sin nx}{(1+2^x)\sin x}\,dx =\displaystyle \int_{-\pi}^{\pi} \dfrac{2^x\sin nx}{(1+2^x)\sin x}\,dx $$

$$I =\frac{1}{2}\int_{-\pi}^{\pi} \dfrac{\sin nx}{\sin x}\,dx = \frac{1}{2}\int_{0}^{2\pi} \dfrac{\sin nx}{\sin x}\,dx $$

The integral is $0$ if $n$ is even and $2\pi$ if it is odd. I will prove it later.

I'm having trouble seeing how any of these integrals are equal?

Where did the extra factor of $\displaystyle \begin{align*} 2^x \end{align*}$ in the integrand come from? How did that all become a factor of $\displaystyle \begin{align*} \frac{1}{2} \end{align*}$? Also how did you change the bounds?
 
ZaidAlyafey said:
$$I=\displaystyle \int_{-\pi}^{\pi} \dfrac{\sin nx}{(1+2^x)\sin x}\,dx =\displaystyle \int_{-\pi}^{\pi} \dfrac{2^x\sin nx}{(1+2^x)\sin x}\,dx $$

$$I =\frac{1}{2}\int_{-\pi}^{\pi} \dfrac{\sin nx}{\sin x}\,dx = \frac{1}{2}\int_{0}^{2\pi} \dfrac{\sin nx}{\sin x}\,dx $$

The integral is $0$ if $n$ is even and $2\pi$ if it is odd. I will prove it later.

Prove It said:
I'm having trouble seeing how any of these integrals are equal?

Where did the extra factor of $\displaystyle \begin{align*} 2^x \end{align*}$ in the integrand come from? How did that all become a factor of $\displaystyle \begin{align*} \frac{1}{2} \end{align*}$? Also how did you change the bounds?

Hi Prove It,

I think given Zaid has mentioned that he would prove how such and such are true, we will wait for his second post for more clarification...(Smile)
 
I just edited my post. I added more clarifications and the proof of the last integral.
 
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