text: ðŸ‘‰ Try now NerdPal! Our new math app on iOS and Android
  
  # 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
  
  ###  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
  
  $\int\frac{1}{3-\frac{1-t^{2}}{1+t^{2}}}\frac{2}{1+t^{2}}dt$
  
  Combine $3-\frac{1-t^{2}}{1+t^{2}}$ in a single fraction
  
  $\int\frac{1}{\frac{-\left(1-t^{2}\right)+3\left(1+t^{2}\right)}{1+t^{2}}}\frac{2}{1+t^{2}}dt$
  
  Simplify the product $-(1-t^{2})$
  
  $\int\frac{1}{\frac{-1+t^{2}+3\left(1+t^{2}\right)}{1+t^{2}}}\frac{2}{1+t^{2}}dt$
  
  Multiply the single term $3$ by each term of the polynomial $\left(1+t^{2}\right)$
  
  $\int\frac{1}{\frac{-1+t^{2}+3+3t^{2}}{1+t^{2}}}\frac{2}{1+t^{2}}dt$
  
  Add the values $-1$ and $3$
  
  $\int\frac{1}{\frac{2+t^{2}+3t^{2}}{1+t^{2}}}\frac{2}{1+t^{2}}dt$
  
  Simplifying
  
  $\int\frac{1}{\frac{2+4t^{2}}{1+t^{2}}}\frac{2}{1+t^{2}}dt$
  
  Divide fractions $\frac{1}{\frac{2+4t^{2}}{1+t^{2}}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$
  
  $\int\frac{1+t^{2}}{2+4t^{2}}\frac{2}{1+t^{2}}dt$
  
  Multiplying fractions $\frac{1+t^{2}}{2+4t^{2}} \times \frac{2}{1+t^{2}}$
  
  $\int\frac{2}{2+4t^{2}}dt$
  
  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 $\sqrt{1}$
  
  $\frac{\sqrt{2}}{2}\cdot \left(\frac{1}{1}\right)\arctan\left(\frac{u}{\sqrt{1}}\right)$
  
  Divide $1$ by $1$
  
  $\frac{\sqrt{2}}{2}\arctan\left(\frac{u}{\sqrt{1}}\right)$
  
  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 $C$
  
  $\frac{\sqrt{2}}{2}\arctan\left(\sqrt{2}\tan\left(\frac{x}{2}\right)\right)+C_0$
  
  $\frac{\sqrt{2}}{2}\arctan\left(\sqrt{2}\tan\left(\frac{x}{2}\right)\right)+C_0$
explanation: The document is a step-by-step solution to an integral problem involving a rational function of sine and cosine. It starts with the integral \(\int\frac{dx}{3-\cos(x)}\) and proceeds to solve it using the Weierstrass substitution. The solution is detailed, showing each step of the substitution and simplification process, and concludes with the final answer.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: ðŸ‘‰ Try now NerdPal! Our new math app on iOS and Android
  
  # 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
  
  ###  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
  
  $\int\frac{1}{3-\frac{1-t^{2}}{1+t^{2}}}\frac{2}{1+t^{2}}dt$
  
  Combine $3-\frac{1-t^{2}}{1+t^{2}}$ in a single fraction
  
  $\int\frac{1}{\frac{-\left(1-t^{2}\right)+3\left(1+t^{2}\right)}{1+t^{2}}}\frac{2}{1+t^{2}}dt$
  
  Simplify the product $-(1-t^{2})$
  
  $\int\frac{1}{\frac{-1+t^{2}+3\left(1+t^{2}\right)}{1+t^{2}}}\frac{2}{1+t^{2}}dt$
  
  Multiply the single term $3$ by each term of the polynomial $\left(1+t^{2}\right)$
  
  $\int\frac{1}{\frac{-1+t^{2}+3+3t^{2}}{1+t^{2}}}\frac{2}{1+t^{2}}dt$
  
  Add the values $-1$ and $3$
  
  $\int\frac{1}{\frac{2+t^{2}+3t^{2}}{1+t^{2}}}\frac{2}{1+t^{2}}dt$
  
  Simplifying
  
  $\int\frac{1}{\frac{2+4t^{2}}{1+t^{2}}}\frac{2}{1+t^{2}}dt$
  
  Divide fractions $\frac{1}{\frac{2+4t^{2}}{1+t^{2}}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$
  
  $\int\frac{1+t^{2}}{2+4t^{2}}\frac{2}{1+t^{2}}dt$
  
  Multiplying fractions $\frac{1+t^{2}}{2+4t^{2}} \times \frac{2}{1+t^{2}}$
  
  $\int\frac{2}{2+4t^{2}}dt$
  
  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 $\sqrt{1}$
  
  $\frac{\sqrt{2}}{2}\cdot \left(\frac{1}{1}\right)\arctan\left(\frac{u}{\sqrt{1}}\right)$
  
  Divide $1$ by $1$
  
  $\frac{\sqrt{2}}{2}\arctan\left(\frac{u}{\sqrt{1}}\right)$
  
  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 $C$
  
  $\frac{\sqrt{2}}{2}\arctan\left(\sqrt{2}\tan\left(\frac{x}{2}\right)\right)+C_0$
  
  $\frac{\sqrt{2}}{2}\arctan\left(\sqrt{2}\tan\left(\frac{x}{2}\right)\right)+C_0$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a step-by-step solution to an integral problem involving a rational function of sine and cosine. It starts with the integral \(\int\frac{dx}{3-\cos(x)}\) and proceeds to solve it using the Weierstrass substitution. The solution is detailed, showing each step of the substitution and simplification process, and concludes with the final answer.
  
  Label: question-and-answer