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Integrals of Rational Functions of Sine and Cosine Calculator

Get detailed solutions to your math problems with our Integrals of Rational Functions of Sine and Cosine step-by-step calculator. Practice your math skills and learn step by step with our math solver. Check out all of our online calculators here.

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Example

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1

Here, we show you a step-by-step solved example of integrals of rational functions of sine and cosine. 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

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

Multiplying fractions $\frac{1}{3-\frac{1-t^{2}}{1+t^{2}}} \times \frac{2}{1+t^{2}}$

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

Multiplying the fraction by $-1$

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

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

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

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

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

Simplify the fraction $\frac{2\left(1+t^{2}\right)}{\left(-1+t^{2}+3\left(1+t^{2}\right)\right)\left(1+t^{2}\right)}$ by $1+t^{2}$

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

Simplifying

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

The integral of a function times a constant ($2$) is equal to the constant times the integral of the function

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

Solve the product $3\left(1+t^{2}\right)$

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

Subtract the values $3$ and $-1$

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

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

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

Simplify the expression

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

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

The power of a product is equal to the product of it's factors raised to the same power

$\sqrt{2}t$
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

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

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

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

The derivative of the linear function is equal to $1$

$\sqrt{2}$
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

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

Any expression multiplied by $1$ is equal to itself

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

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

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

After replacing everything and simplifying, the integral results in

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

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

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

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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