# Evaluate $\int\frac {\sin 4x }{\sin x}\ dx$

Evaluate $$\int \frac{\sin 4x}{\sin x} dx$$

Attempt: I've tried to use the double angle formulas and get it all into one identity, which came out as: $$\int \left( 8\cos^3x - 4\cos x \right) dx$$ But I'm not sure if this is the right way to go about it. Any help would be much appreciated!

• It is a good way. Now note that $\cos^3 x=(\cos x)(1-\sin^2 x)$. Commented Oct 23, 2015 at 17:50

$$\frac{\sin(4x)}{\sin(x)}=U_3(\cos(x)) = 8\cos^3(x)-4\cos(x) = 2T_3(\cos x)+2T_1(\cos(x))$$ hence: $$\frac{\sin(4x)}{\sin(x)} = 2\cos(3x)+2\cos(x)$$ implies: $$\int\frac{\sin(4x)}{\sin(x)}\,dx = \frac{2}{3}\sin(3x)+2\sin(x)+C.$$

• Nice, but it would be worth to mention $T_n$ and $U_n$ are Chebyshev polynomials of the first and second kind, respectively ;-) Commented Oct 23, 2015 at 17:51

Notice, $$\int\frac{\sin 4x}{\sin x}\ dx$$ $$=\int\frac{2\sin 2x\cos 2x}{\sin x}\ dx$$ $$=\int\frac{2(2\sin x\cos x) (1-2\sin^2x)}{\sin x}\ dx$$ $$=\int4(\cos x -2\sin^2x\cos x)\ dx$$ $$=4\int\cos x -8\int \sin^2x\cos x\ dx$$ $$=4\int\cos x -8\int \sin^2x\ d(\sin x)$$ $$=\color{blue}{4\sin x-\frac{8\sin^3 x}{3}+C}$$

We have $$\frac{\sin(4x)}{\sin x}=\frac{2\sin(2x)\cos(2x)}{\sin x}=\frac{4\sin x\cos x\cos(2x)}{\sin x}$$ $$=4\cos x\cos(2x)=4\cos x(1-2\sin^2x)$$

Now set $\sin x=t$.

Here is another solution based on Dirichlet kernel

Lemma: $\sum _{k=-n} ^{n} e^{ikx} = 1 + 2 \sum _{k=1} ^{n} cos(kx) = \frac{sin((n+1)/2)}{sin(x/2)}$

Proof:

It's direct application for series sum $\sum _{k=-n} ^{n} x^k$, and Euler celebrated identity $e^{ix} = cos(x) + i sin(x)$, see wikipedia entry for full proof see wikipedia's entry

Now: put $x = \frac{u}{2} \Rightarrow dx = \frac{1}{2}du$

$\frac{1}{2} \int \frac{\sin 2u}{\sin \frac{u}{2}} dx = \int (1 +2 \sum _{k=1} ^{1} cos(ku)) dx = u - 2 sin(u)$

Note: This method works for any positive integer $n$ in $\int \frac{sin(2nx)}{sin(x)}$

We can put the complex identity $$\sin \alpha := \frac{\exp(i \alpha) - \exp(-i \alpha)}{2i}$$ to efficient use here.

Taking $\alpha = 4 x$ gives $$\sin 4x = \frac{\exp(4 i x) - \exp(-4 i x)}{2i},$$ We can use a difference-of-squares factorization to write the numerator as \begin{align*} \exp(4 i x) - \exp(-4 i x) &= [\exp(2 i x) + \exp(-2 i x)][\exp(2 i x - \exp(2 i x)] \\ &= [\exp(2 i x) + \exp(-2 i x)][\exp (i x) + \exp(- i x)][\exp (i x) - \exp(- i x)] \end{align*} Now, substituting using the complex identity for $\alpha = x$ gives \begin{align*} \sin 4x = [\exp(2 i x) + \exp(-2 i x)][\exp (i x) + \exp(- i x)] \cdot \frac{\exp (i x) - \exp(- i x)}{2 i} = [\exp(2 i x) + \exp(-2 i x)][\exp (i x) + \exp(- i x)] \sin x. \end{align*} So, expanding and using the corresponding cosine identity $$\cos \alpha = \frac{\exp(i \alpha) + \exp(-i \alpha)}{2}$$ gives \begin{align*} \frac{\sin 4x}{\sin x} &= [\exp(2 i x) + \exp(-2 i x)][\exp (i x) + \exp(- i x)]\\ &= \exp(3 i x) + \exp(-3 i x) + \exp(i x) + \exp(-i x) \\ &= 2 (\cos 3x - \cos x) . \end{align*} In this form, we can compute the antiderivative quickly.

$$\int\;\frac {sin(4x)}{sin(x)}dx＝\int\;\frac {2sin(2x)cos(2x)}{sin(x)}dx＝\int4cos(x)cos(2x)dx\\~\\=4\int\;cos(x)(1-2sin^2(x))dx$$ let u = sin(x) and du = -cos(x) dx and therfore : $$＝－4\int1-2u^2du＝-4u＋\frac {8}{3}u^3＋C＝-4sin(x)＋\frac {8}{3}sin(x)^3＋C$$

$$\int \left( 8\cos^3x - 4\cos x \right) dx=\int \left( 8\cos x (1-\sin^2 x) - 4\cos x \right) dx = \int \left( -8\cos x \sin^2 x + 4\cos x \right) dx=-8\int \cos x \sin^2 x dx + 4\int cos x dx = \frac{-8\sin^3 x}{3} +4\sin x + C$$