Integrals of Rational Functions of Sine and Cosine Calculator

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Difficult Problems

1

Solved example of integrals of rational functions of sine and cosine

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

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

Combine $3-\left(\frac{1-t^{2}}{1+t^{2}}\right)$ in a single fraction

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

Divide fractions $\frac{1}{\frac{-\left(1-t^{2}\right)+3\left(1+t^{2}\right)}{1+t^{2}}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$

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

Solve the product $-(1-t^{2})$

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

Multiplying polynomials $3$ and $1+t^{2}$

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

Multiplying fractions $\frac{1+t^{2}}{2+4t^{2}} \times \frac{2}{1+t^{2}}$

$\int\frac{2\left(1+t^{2}\right)}{\left(2+4t^{2}\right)\left(1+t^{2}\right)}dt$

Simplify the fraction $\frac{2\left(1+t^{2}\right)}{\left(2+4t^{2}\right)\left(1+t^{2}\right)}$ by $1+t^{2}$

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

Factor the denominator by $2$

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

Cancel the fraction's common factor $2$

$\int\frac{1}{1+2t^{2}}dt$
5

Simplifying

$\int\frac{1}{1+2t^{2}}dt$
6

Solve the integral applying the substitution $u^2=2t^{2}$

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

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

Calculate the power $\sqrt{1}$

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

Any expression divided by one ($1$) is equal to that same expression

$\frac{\sqrt{2}}{2}\arctan\left(u\right)$
7

Solve the integral applying the formula $\displaystyle\int\frac{x'}{x^2+a^2}dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)$

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

$\frac{\sqrt{2}}{2}\arctan\left(\frac{2}{\sqrt{2}}t\right)$
8

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

$\frac{\sqrt{2}}{2}\arctan\left(\frac{2}{\sqrt{2}}t\right)$
9

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

$\frac{\sqrt{2}}{2}\arctan\left(\frac{2}{\sqrt{2}}\tan\left(\frac{x}{2}\right)\right)$
10

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

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

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