👉 Try now NerdPal! Our new math app on iOS and Android
  1. calculators
  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.

Go!
Math mode
Text mode
Go!
1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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

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

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

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

Multiplying fractions $\frac{1}{\frac{-\left(1-t^{2}\right)+3\left(1+t^{2}\right)}{1+t^{2}}} \times \frac{2}{1+t^{2}}$

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

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

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

Multiply the single term $3$ by each term of the polynomial $\left(1+t^{2}\right)$

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

Add the values $-1$ and $3$

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

Simplifying

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

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

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

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

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

$\sqrt{2}t$
6

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 the linear function times a constant, is equal to the constant

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

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

$\sqrt{2}$
7

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

$du=\sqrt{2}dt$
8

Isolate $dt$ in the previous equation

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

Divide $1$ by $\sqrt{2}$

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

Divide $\frac{\sqrt{2}}{2}$ by $1$

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

Any expression multiplied by $1$ is equal to itself

$\frac{\sqrt{2}}{2}\int\frac{1}{1+u^2}du$
9

After replacing everything and simplifying, the integral results in

$\frac{\sqrt{2}}{2}\int\frac{1}{1+u^2}du$
10

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

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

Calculate the power $\sqrt{1}$

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

Simplify the fraction $\frac{1}{1}$ by $1$

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

Multiply $\frac{\sqrt{2}}{2}$ times $1$

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

Calculate the square root of $1$

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

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

$\frac{\sqrt{2}}{2}\arctan\left(u\right)$
11

Simplify the expression inside the integral

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

$\frac{\sqrt{2}}{2}\arctan\left(\sqrt{2}t\right)$
12

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

$\frac{\sqrt{2}}{2}\arctan\left(\sqrt{2}t\right)$
13

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

$\frac{\sqrt{2}}{2}\arctan\left(\sqrt{2}\tan\left(\frac{x}{2}\right)\right)$
14

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

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

Final Answer

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

Struggling with math?

Access detailed step by step solutions to thousands of problems, growing every day!