Calculus

Integration of Functions

Integration of Rational Expressions of Trigonometric Functions

Page 1
Problems 1-2
Page 2
Problems 3-7

Any rational expression of trigonometric functions can be always reduced to integrating a rational function by making the universal trigonometric substitution \(x = 2\arctan t\) (or \(t = \tan {\large\frac{x}{2}\normalsize}\)).

The following trigonometric formulas are used to transform rational expressions of \(\sin x,\) \(\cos x,\) \(\tan x,\) \(\cot x,\) \(\sec x\) and \(\csc x\) into rational functions of \(t:\)

\(\sin x = {\large\frac{{2\tan \frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}}\normalsize} = {\large\frac{{2t}}{{1 + {t^2}}}\normalsize}\) \(\cos x = {\large\frac{{1 – {{\tan }^2}\frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}}\normalsize} = {\large\frac{{1 – {t^2}}}{{1 + {t^2}}}\normalsize}\)
\(\tan x = {\large\frac{{2\tan \frac{x}{2}}}{{1 – {{\tan }^2}\frac{x}{2}}}\normalsize} = {\large\frac{{2t}}{{1 – {t^2}}}\normalsize}\) \(\cot x = {\large\frac{{1 – {{\tan }^2}\frac{x}{2}}}{{2\tan \frac{x}{2}}}\normalsize} = {\large\frac{{1 – {t^2}}}{{2t}}\normalsize}\)
\(\sec x = {\large\frac{{1 + {{\tan }^2}\frac{x}{2}}}{{2\tan \frac{x}{2}}}\normalsize} = {\large\frac{{1 + {t^2}}}{{2t}}\normalsize}\) \(\csc x = {\large\frac{{1 + {{\tan }^2}\frac{x}{2}}}{{1 – {{\tan }^2}\frac{x}{2}}}\normalsize} = {\large\frac{{1 + {t^2}}}{{1 – {t^2}}}\normalsize}\)
\(\sin x = {\large\frac{{2\tan \frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}}\normalsize} = {\large\frac{{2t}}{{1 + {t^2}}}\normalsize}\)
\(\cos x = {\large\frac{{1 – {{\tan }^2}\frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}}\normalsize} = {\large\frac{{1 – {t^2}}}{{1 + {t^2}}}\normalsize}\)
\(\tan x = {\large\frac{{2\tan \frac{x}{2}}}{{1 – {{\tan }^2}\frac{x}{2}}}\normalsize} = {\large\frac{{2t}}{{1 – {t^2}}}\normalsize}\)
\(\cot x = {\large\frac{{1 – {{\tan }^2}\frac{x}{2}}}{{2\tan \frac{x}{2}}}\normalsize} = {\large\frac{{1 – {t^2}}}{{2t}}\normalsize}\)
\(\sec x = {\large\frac{{1 + {{\tan }^2}\frac{x}{2}}}{{2\tan \frac{x}{2}}}\normalsize} = {\large\frac{{1 + {t^2}}}{{2t}}\normalsize}\)
\(\csc x = {\large\frac{{1 + {{\tan }^2}\frac{x}{2}}}{{1 – {{\tan }^2}\frac{x}{2}}}\normalsize} = {\large\frac{{1 + {t^2}}}{{1 – {t^2}}}\normalsize}\)

To calculate an integral of the form \({\large\int\normalsize} {R\left( {\sin x} \right)\cos x\,dx} ,\) where \(R\) is a rational function, use the substitution \(t = \sin x.\)

Similarly, to calculate an integral of the form \({\large\int\normalsize} {R\left( {\cos x} \right)\sin x\,dx} ,\) where \(R\) is a rational function, use the substitution \(t = \cos x.\)

If an integrand is a function of only \(\tan x,\) the substitution \(t = \tan x\) converts this integral into integral of a rational function.

To calculate an integral of the form \({\large\int\normalsize} {R\left( {\sin x} \right)\cos x\,dx} ,\) where both functions \(\sin x\) and \(\cos x\) have even powers, use the substitution \(t = \tan x\) and the formulas
\[
{{{\cos ^2}x = \frac{1}{{1 + {{\tan }^2}x}} }={ \frac{1}{{1 + {t^2}}},\;\;\;}}\kern0pt
{{{\sin ^2}x = \frac{{{{\tan }^2}x}}{{1 + {{\tan }^2}x}} }={ \frac{{{t^2}}}{{1 + {t^2}}}} }
\]

Solved Problems

Click on problem description to see solution.

 Example 1

Evaluate the integral \({\large\int\normalsize} {\large\frac{{dx}}{{1 + \sin x}}\normalsize}.\)

 Example 2

Calculate the integral \({\large\int\normalsize} {\large\frac{{dx}}{{1 + \cos \frac{x}{2}}}\normalsize} .\)

 Example 3

Find the integral \({\large\int\normalsize} {\large\frac{{dx}}{{\sin x + \cos x}}\normalsize}. \)

 Example 4

Evaluate \({\large\int\normalsize} {\large\frac{{dx}}{{\sec x + 1}}\normalsize} .\)

 Example 5

Calculate the integral \({\large\int\normalsize} {\large\frac{{dx}}{{{{\sin }^4}x + {{\cos }^4}x}}\normalsize}.\)

 Example 6

Calculate the integral \({\large\int\normalsize} {\large\frac{{dx}}{{a\sin x + b\cos x}}\normalsize}.\)

 Example 7

Find the integral \({\large\int\normalsize} {\large\frac{{dx}}{{1 + \tan x}}\normalsize} .\)

Example 1.

Evaluate the integral \({\large\int\normalsize} {\large\frac{{dx}}{{1 + \sin x}}\normalsize}.\)

Solution.

We use the universal trigonometric substitution:
\[
{t = \tan \frac{x}{2},\;\;}\Rightarrow
{{x = 2\arctan t,\;\;\;}\kern-0.3pt{dx = \frac{{2dt}}{{1 + {t^2}}}.}}
\] Since \(\sin x = {\large\frac{{2t}}{{1 + {t^2}}}\normalsize},\) we have
\[
{\int {\frac{{dx}}{{1 + \sin x}}} }
= {\int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{1 + \frac{{2t}}{{1 + {t^2}}}}}} }
= {\int {\frac{{dt}}{{1 + {t^2} + 2t}}} }
= {\int {\frac{{dt}}{{{{\left( {t + 1} \right)}^2}}}} }
= { – \frac{2}{{t + 1}} + C }
= { – \frac{2}{{\tan \frac{x}{2} + 1}} + C.}
\]

Example 2.

Calculate the integral \({\large\int\normalsize} {\large\frac{{dx}}{{1 + \cos \frac{x}{2}}}\normalsize} .\)

Solution.

Make the universal trigonometric substitution:
\[
{t = \tan \frac{x}{4},\;\;}\Rightarrow
{d\left( {\frac{x}{2}} \right) = \frac{{2dt}}{{1 + {t^2}}},\;\;}\Rightarrow
{\cos \frac{x}{2} = \frac{{1 – {t^2}}}{{1 + {t^2}}}.}
\] Then the integral becomes
\[\require{cancel}
{\int {\frac{{dx}}{{1 + \cos \frac{x}{2}}}} }
={ \int {\frac{{d\left( {\frac{x}{2}} \right)}}{{1 + \cos \frac{x}{2}}}} }
={ 2\int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{1 + \frac{{1 – {t^2}}}{{1 + {t^2}}}}}} }
={ 4\int {\frac{{dt}}{{1 + \cancel{t^2} + 1 – \cancel{t^2}}}} }
={ 2\int {dt} }
={ 2t + C }
={ 2\tan \frac{x}{4} + C.}
\]

Page 1
Problems 1-2
Page 2
Problems 3-7