Calculus

Integration of Functions

Integration of Functions Logo

Integration of Rational Expressions of Trigonometric Functions

  • 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 \(t:\)

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

    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 a problem to see the 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