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  2. Integrals Of Rational Functions Of Sine And Cosine

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 finding the derivative of 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 exercise

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

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