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Intégrales des fonctions rationnelles du sinus et du cosinus Calculator

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1

Here, we show you a step-by-step solved example of intégrales des fonctions rationnelles du sinus et du cosinus. This solution was automatically generated by our smart calculator:

$\int\frac{dx}{3-cos\left(x\right)}$
2

We can solve the integral $\int\frac{1}{3-\cos\left(x\right)}dx$ by applying the method Weierstrass substitution (also known as tangent half-angle substitution) which converts an integral of trigonometric functions into a rational function of $t$ by setting the substitution

$t=\tan\left(\frac{x}{2}\right)$
3

Hence

$\sin x=\frac{2t}{1+t^{2}},\:\cos x=\frac{1-t^{2}}{1+t^{2}},\:\mathrm{and}\:\:dx=\frac{2}{1+t^{2}}dt$
4

Substituting in the original integral we get

$\int\frac{1}{3-\frac{1-t^{2}}{1+t^{2}}}\frac{2}{1+t^{2}}dt$

Apply the formula: $\frac{a}{b}\frac{c}{f}$$=\frac{ac}{bf}$, where $a=1$, $b=3-\frac{1-t^{2}}{1+t^{2}}$, $c=2$, $a/b=\frac{1}{3-\frac{1-t^{2}}{1+t^{2}}}$, $f=1+t^{2}$, $c/f=\frac{2}{1+t^{2}}$ and $a/bc/f=\frac{1}{3-\frac{1-t^{2}}{1+t^{2}}}\frac{2}{1+t^{2}}$

$\int\frac{2}{\left(3-\frac{1-t^{2}}{1+t^{2}}\right)\left(1+t^{2}\right)}dt$

Apply the formula: $-\frac{b}{c}$$=\frac{expand\left(-b\right)}{c}$, where $b=1-t^{2}$ and $c=1+t^{2}$

$\int\frac{2}{\left(3+\frac{-1+t^{2}}{1+t^{2}}\right)\left(1+t^{2}\right)}dt$

Apply the formula: $a+\frac{b}{c}$$=\frac{b+ac}{c}$, where $a=3$, $b=-1+t^{2}$, $c=1+t^{2}$, $a+b/c=3+\frac{-1+t^{2}}{1+t^{2}}$ and $b/c=\frac{-1+t^{2}}{1+t^{2}}$

$\int\frac{2}{\frac{-1+t^{2}+3\left(1+t^{2}\right)}{1+t^{2}}\left(1+t^{2}\right)}dt$

Apply the formula: $\frac{a}{\frac{b}{c}}$$=\frac{ac}{b}$, where $a=2$, $b=-1+t^{2}+3\left(1+t^{2}\right)$, $c=1+t^{2}$, $a/b/c=\frac{2}{\frac{-1+t^{2}+3\left(1+t^{2}\right)}{1+t^{2}}\left(1+t^{2}\right)}$ and $b/c=\frac{-1+t^{2}+3\left(1+t^{2}\right)}{1+t^{2}}$

$\int\frac{2\left(1+t^{2}\right)}{\left(-1+t^{2}+3\left(1+t^{2}\right)\right)\left(1+t^{2}\right)}dt$

Apply the formula: $\frac{a}{a}$$=1$, where $a=1+t^{2}$ and $a/a=\frac{2\left(1+t^{2}\right)}{\left(-1+t^{2}+3\left(1+t^{2}\right)\right)\left(1+t^{2}\right)}$

$\int\frac{2}{-1+t^{2}+3\left(1+t^{2}\right)}dt$
5

Simplifying

$\int\frac{2}{-1+t^{2}+3\left(1+t^{2}\right)}dt$
6

Apply the formula: $\int\frac{n}{a+b}dx$$=n\int\frac{1}{a+b}dx$, where $a=-1$, $b=t^{2}+3\left(1+t^{2}\right)$ and $n=2$

$2\int\frac{1}{-1+t^{2}+3\left(1+t^{2}\right)}dt$
7

Apply the formula: $x\left(a+b\right)$$=xa+xb$, where $a=1$, $b=t^{2}$, $x=3$ and $a+b=1+t^{2}$

$2\int\frac{1}{t^{2}+3+3t^{2}-1}dt$

Apply the formula: $a+b$$=a+b$, where $a=3$, $b=-1$ and $a+b=t^{2}+3+3t^{2}-1$

$2\int\frac{1}{t^{2}+2+3t^{2}}dt$

Combining like terms $t^{2}$ and $3t^{2}$

$2\int\frac{1}{4t^{2}+2}dt$
8

Simplify the expression

$2\int\frac{1}{4t^{2}+2}dt$

$\sqrt{2t^{2}}$

Apply the formula: $\left(ab\right)^n$$=a^nb^n$, where $a=2$, $b=t^{2}$ and $n=\frac{1}{2}$

$\sqrt{2}t$
9

Solve the integral applying the substitution $u^2=2t^{2}$. Then, take the square root of both sides, simplifying we have

$u=\sqrt{2}t$

Differentiate both sides of the equation $u=\sqrt{2}t$

$du=\frac{d}{dt}\left(\sqrt{2}t\right)$

Find the derivative

$\frac{d}{dt}\left(\sqrt{2}t\right)$

Apply the formula: $\frac{d}{dx}\left(cx\right)$$=c\frac{d}{dx}\left(x\right)$

$\sqrt{2}\frac{d}{dt}\left(t\right)$

Apply the formula: $\frac{d}{dx}\left(x\right)$$=1$, where $x=t$

$\sqrt{2}$
10

Now, in order to rewrite $dt$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=\sqrt{2}dt$
11

Isolate $dt$ in the previous equation

$\frac{du}{\sqrt{2}}=dt$

Apply the formula: $1x$$=x$, where $x=u^2$

$2\cdot \left(\frac{\frac{1}{\sqrt{2}}}{2}\right)\int\frac{1}{u^2+1}du$

Apply the formula: $a\frac{x}{b}$$=\frac{a}{b}x$, where $a=2$, $b=2$, $ax/b=2\cdot \left(\frac{\frac{1}{\sqrt{2}}}{2}\right)\int\frac{1}{u^2+1}du$, $x=\frac{1}{\sqrt{2}}$ and $x/b=\frac{\frac{1}{\sqrt{2}}}{2}$

$\frac{1}{\sqrt{2}}\int\frac{1}{u^2+1}du$
12

After replacing everything and simplifying, the integral results in

$\frac{1}{\sqrt{2}}\int\frac{1}{u^2+1}du$
13

Apply the formula: $\int\frac{n}{x^2+b}dx$$=\frac{n}{\sqrt{b}}\arctan\left(\frac{x}{\sqrt{b}}\right)+C$, where $b=1$, $x=u$ and $n=1$

$\frac{1}{\sqrt{2}}\arctan\left(u\right)$

Replace $u$ with the value that we assigned to it in the beginning: $\sqrt{2}t$

$\frac{1}{\sqrt{2}}\arctan\left(\sqrt{2}t\right)$
14

Replace $u$ with the value that we assigned to it in the beginning: $\sqrt{2}t$

$\frac{1}{\sqrt{2}}\arctan\left(\sqrt{2}t\right)$
15

Replace $t$ with the value that we assigned to it in the beginning: $\tan\left(\frac{x}{2}\right)$

$\frac{1}{\sqrt{2}}\arctan\left(\sqrt{2}\tan\left(\frac{x}{2}\right)\right)$
16

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$\frac{1}{\sqrt{2}}\arctan\left(\sqrt{2}\tan\left(\frac{x}{2}\right)\right)+C_0$

Final answer to the problem

$\frac{1}{\sqrt{2}}\arctan\left(\sqrt{2}\tan\left(\frac{x}{2}\right)\right)+C_0$

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