0
$\begingroup$

I struggle solving the following.

$f''+9f=\cot(3x)$

Homogeneous:

With the Ansatz $f=e^{\lambda x}$ we find

$$P(\lambda)=\lambda^2+9=0 \quad \Rightarrow \quad \lambda_{1,2}=\pm3i$$

so we get

$$y_h(x)=A\cos(3x)+B\sin(3x)$$

Particular:

We use variation of constants and we consider $A,B$ as two functions of x: $A(x), B(x)$. With the Basis $\{\cos(3x), \sin(3x)\}$ we find:

$$\begin{pmatrix}\cos(3x)& \sin(3x) \\ -3\sin(3x) & 3\cos(3x)\end{pmatrix}\cdot\begin{pmatrix}A'(x)\\B'(x)\end{pmatrix}=\begin{pmatrix}0\\ \frac{\cos(3x)}{\sin(3x)}\end{pmatrix}$$

We solve it and get

$$A'(x)=-\frac{1}{3}\cos(3x), \quad B'(X)=\frac{\cos^2(x)}{\sin(3x)}$$

We integrate it:

$$A(x)=\int A'(x) dx = \frac{1}{3}\sin(3x)$$

$B(x)=\int B'(x) dx = \frac{1}{3}\int\frac{\cos^2(3x)}{\sin(3x)}dx=\frac{1}{3}\int\frac{1-\sin^2(3x)}{\sin(3x)}=\frac{1}{3}\int\frac{1}{\sin(3x)}dx-\frac{1}{3}\int \sin(3x)dx$

Now I have problems solving $\int\frac{1}{\sin(3x)}dx$

Any hints? :)

$\endgroup$
  • 3
    $\begingroup$ Start by letting $3x=t$ , then to integrate $\int \frac{dt}{\sin t}$ multiply it by $\frac{\sin t}{\sin t}$. You have to think though how to rewrite $\sin^2 t$ from the denominator in order to solve the integral :) $\endgroup$ – LeBlanc Jan 9 at 17:36
  • 1
    $\begingroup$ Answer is $\frac{1}{3}ln|\arctan(\frac{3x}{2})|+c$. The step by step solution is here: symbolab.com/solver/integral-calculator/… $\endgroup$ – ersh Jan 9 at 17:43
  • $\begingroup$ Thanks, gonna try it. $\endgroup$ – xotix Jan 9 at 17:46
  • $\begingroup$ This is similar to the secant integral. There are many forms of the solution, a common one is $\frac13 \ln|\csc 3x + \cot 3x|$ $\endgroup$ – Dylan Jan 9 at 18:08
  • 1
    $\begingroup$ @Dylan watch your signs, it's $\frac{1}{3}\ln|\csc 3x\color{red}{-}\cot 3x|+C$. $\endgroup$ – Oscar Lanzi Jan 10 at 1:22
3
$\begingroup$

\begin{equation} I = \int \frac{1}{\sin\left(3x\right)}\:dx \end{equation}

Let $u = 3x$ to yield:

\begin{equation} I = \int \frac{1}{\sin\left(3x\right)}dx = \int \frac{1}{\sin\left(u\right)}\cdot\frac{1}{3}\:du = \frac{1}{3}\int \operatorname{cosec}(u) \:du \end{equation}

There are a variety of ways to approach this integral. One method is covered here. You will observe that method requires knowledge of a trigonometric derivative identity. Here I will employ a method that can be used to solve integrals of rational expressions of trigonometric functions. This method is known as the Half Tangent (aka Weierstrass) Substitution. Thus we let $t = \tan\left(\frac{u}{2} \right)$ to yield:

\begin{align} I &= \frac{1}{3}\int \operatorname{cosec}(u) \:du = \frac{1}{3}\int \frac{1}{\sin(u)}\:du \\ &= \frac{1}{3}\int \frac{1}{\frac{2t}{1 + t^2}}\frac{2}{1 + t^2}\:dt = \frac{1}{3} \int \frac{1}{t}\:dt = \frac{1}{3}\ln\left|t \right| + C \\ &= \frac{1}{3}\ln\left|\:\tan\left(\frac{u}{2}\right)\: \right| + C = \frac{1}{3}\ln\left|\:\tan\left(\frac{3x}{2}\right) \: \right| + C \end{align}

Where $C$ is the constant of integration.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.