Calculus

Integration of Functions

Integration of Functions Logo

Weierstrass Substitution

  • The Weierstrass substitution, named after German mathematician Karl Weierstrass \(\left({1815 – 1897}\right),\) is used for converting rational expressions of trigonometric functions into algebraic rational functions, which may be easier to integrate.

    This method of integration is also called the tangent half-angle substitution as it implies the following half-angle identities:

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

    where \(t = \tan \large{\frac{x}{2}}\normalsize\) or \(x = 2\arctan t.\)

    The differential \(dx\) is determined as follows:

    \[{dx = d\left( {2\arctan t} \right) }={ \frac{{2dt}}{{1 + {t^2}}}.}\]

    Any rational expression of trigonometric functions can be always reduced to integrating a rational function by making the Weierstrass substitution.

    The Weierstrass substitution is very useful for integrals involving a simple rational expression in \(\sin x\) and/or \(\cos x\) in the denominator.

    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

    Evaluate the integral \(\int {\large{\frac{{dx}}{{3 – 2\sin x}}}\normalsize}.\)

    Example 3

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

    Example 4

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

    Example 5

    Compute the integral \(\int {\large{\frac{{dx}}{{4 + 5\cos \frac{x}{2}}}}\normalsize}.\)

    Example 6

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

    Example 7

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

    Example 8

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

    Example 9

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

    Example 10

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

    Example 11

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

    Example 12

    Find the integral \(\int {\large{\frac{{dx}}{{3\sin x + 4\cos x}}}\normalsize}.\)

    Example 13

    Find the integral \(\int {\large{\frac{{dx}}{{5\sin x + 12\cos x}}}\normalsize}.\)

    Example 14

    Compute the integral \(\int {\large{\frac{{dx}}{{5\sin x + 2\cos x + 2}}}\normalsize}.\)

    Example 15

    Find the integral \(\int {\large{\frac{{dx}}{{2\sin x – \cos x + 5}}}\normalsize}.\)

    Example 16

    Find the integral \(\int {\large{\frac{{2dx}}{{4\sin x – 3\cos x + 5}}}\normalsize}.\)

    Example 17

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

    Example 18

    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.

    Evaluate the integral \(\int {\large{\frac{{dx}}{{3 – 2\sin x}}}\normalsize}.\)

    Solution.

    Using the Weierstrass substitution

    \[{x = \arctan t,\;\;}\kern0pt{\sin x = \frac{{2t}}{{1 + {t^2}}},\;\;}\kern0pt{dx = \frac{{2dt}}{{1 + {t^2}}},}\]

    we can rewrite the integral in the form

    \[{I = \int {\frac{{dx}}{{3 – 2\sin x}}} }={ \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{3 – 2 \cdot \frac{{2t}}{{1 + {t^2}}}}}} }={ \int {\frac{{2dt}}{{3 + 3{t^2} – 4t}}} }={ \int {\frac{{2dt}}{{3\left( {{t^2} – \frac{4}{3}t + 1} \right)}}} }={ \frac{2}{3}\int {\frac{{dt}}{{{t^2} – \frac{4}{3}t + 1}}} .}\]

    Complete the square in the denominator:

    \[{{t^2} – \frac{4}{3}t + 1 }={ {t^2} – \frac{4}{3}t + {\left( {\frac{2}{3}} \right)^2} – {\left( {\frac{2}{3}} \right)^2} + 1 }={ {\left( {t – \frac{2}{3}} \right)^2} – \frac{4}{9} + 1 }={ {\left( {t – \frac{2}{3}} \right)^2} + \frac{5}{9} }={ {\left( {t – \frac{2}{3}} \right)^2} + {\left( {\frac{{\sqrt 5 }}{3}} \right)^2}.}\]

    Changing \(u = t – \large{\frac{2}{3}}\normalsize,\) \(du = dt\) gives the final answer:

    \[{I = \frac{2}{3}\int {\frac{{dt}}{{{{\left( {t – \frac{2}{3}} \right)}^2} + {{\left( {\frac{{\sqrt 5 }}{3}} \right)}^2}}}} }={ \frac{2}{3}\int {\frac{{du}}{{{u^2} + {{\left( {\frac{{\sqrt 5 }}{3}} \right)}^2}}}} }={ \frac{2}{3} \cdot \frac{1}{{\frac{{\sqrt 5 }}{3}}}\arctan \frac{u}{{\frac{{\sqrt 5 }}{3}}} + C }={ \frac{2}{{\sqrt 5 }}\arctan \frac{{3\left( {t – \frac{2}{3}} \right)}}{{\sqrt 5 }} + C }={ \frac{2}{{\sqrt 5 }}\arctan \frac{{3t – 2}}{{\sqrt 5 }} + C }={ \frac{2}{{\sqrt 5 }}\arctan \left( {\frac{{3\tan \frac{x}{2} – 2}}{{\sqrt 5 }}} \right) }+{ C.}\]

    Example 3.

    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

    \[\require{cancel}
    {\int {\frac{{dx}}{{1 + \cos \frac{x}{2}}}} }
    ={ \int {\frac{{d\left( {\frac{x}{2}} \right)}}{{1 + \cos \frac{x}{2}}}} }
    ={ 2\int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{1 + \frac{{1 – {t^2}}}{{1 + {t^2}}}}}} }
    ={ 4\int {\frac{{dt}}{{1 + \cancel{t^2} + 1 – \cancel{t^2}}}} }
    ={ 2\int {dt} }
    ={ 2t + C }
    ={ 2\tan \frac{x}{4} + C.}
    \]

    Example 4.

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

    Solution.

    Using the substitution

    \[{t = \tan x,}\;\; \Rightarrow {x = \arctan t,}\;\; \Rightarrow {dx = \frac{{dt}}{{1 + {t^2}}},}\;\; \Rightarrow {\cos 2x = \frac{{1 – {t^2}}}{{1 + {t^2}}},}\]

    we can easily find the integral:

    \[{\require{cancel}\int {\frac{{dx}}{{1 + \cos 2x}}} }={ \int {\frac{{\frac{{dt}}{{1 + {t^2}}}}}{{1 + \frac{{1 – {t^2}}}{{1 + {t^2}}}}}} }={ \int {\frac{{dt}}{{1 + \cancel{t^2} + 1 – \cancel{t^2}}}} }={ \int {\frac{{dt}}{2}} }={ \frac{t}{2} + C }={ \frac{1}{2}\tan x + C.}\]

    Example 5.

    Compute the integral \(\int {\large{\frac{{dx}}{{4 + 5\cos \frac{x}{2}}}}\normalsize}.\)

    Solution.

    To simplify the integral, we use the Weierstrass substitution:

    \[{t = \tan \frac{x}{4},}\;\; \Rightarrow {x = 4\arctan t,\;\;}\kern0pt{dx = \frac{{4dt}}{{1 + {t^2}}},\;\;}\kern0pt{\cos \frac{x}{2} = \frac{{1 – {t^2}}}{{1 + {t^2}}}.}\]

    The integral becomes

    \[{\int {\frac{{dx}}{{4 + 5\cos \frac{x}{2}}}} }={ \int {\frac{{\frac{{4dt}}{{1 + {t^2}}}}}{{4 + 5 \cdot \frac{{1 – {t^2}}}{{1 + {t^2}}}}}} }={ \int {\frac{{4dt}}{{4\left( {1 + {t^2}} \right) + 5\left( {1 – {t^2}} \right)}}} }={ 4\int {\frac{{dt}}{{4 + 4{t^2} + 5 – 5{t^2}}}} }={ 4\int {\frac{{dt}}{{{3^2} – {t^2}}}} }={ 4 \cdot \frac{1}{{2 \cdot 3}}\ln \left| {\frac{{3 + t}}{{3 – t}}} \right| + C }={ \frac{2}{3}\ln \left| {\frac{{3 + \tan \frac{x}{4}}}{{3 – \tan \frac{x}{4}}}} \right| + C.}\]

    Example 6.

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

    Solution.

    As in the previous examples, we will 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 \(\sin x = {\large\frac{{2t}}{{1 + {t^2}}\normalsize}},\) \(\cos x = {\large\frac{{1 – {t^2}}}{{1 + {t^2}}\normalsize}},\) we can write:

    \[
    {\int {\frac{{dx}}{{\sin x + \cos x}}} }
    ={ \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{\frac{{2t}}{{1 + {t^2}}} + \frac{{1 – {t^2}}}{{1 + {t^2}}}}}} }
    ={ \int {\frac{{2dt}}{{2t + 1 – {t^2}}}} }
    ={ 2\int {\frac{{dt}}{{1 – \left( {{t^2} – 2t} \right)}}} }
    ={ 2\int {\frac{{dt}}{{1 – \left( {{t^2} – 2t + 1 – 1} \right)}}} }
    ={ 2\int {\frac{{dt}}{{2 – {{\left( {t – 1} \right)}^2}}}} }
    ={ 2\int {\frac{{d\left( {t – 1} \right)}}{{{{\left( {\sqrt 2 } \right)}^2} – {{\left( {t – 1} \right)}^2}}}} }
    ={{ 2 \cdot}\kern0pt{ \frac{1}{{2\sqrt 2 }}\ln \left| {\frac{{\sqrt 2 + \left( {t – 1} \right)}}{{\sqrt 2 – \left( {t – 1} \right)}}} \right| }+{ C }}
    ={{ \frac{1}{{\sqrt 2 }}\ln \left| {\frac{{\sqrt 2 – 1 + \tan \frac{x}{2}}}{{\sqrt 2 + 1 – \tan \frac{x}{2}}}} \right| }+{ C.}}
    \]

    Example 7.

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

    Solution.

    Making the \({\tan \large{\frac{x}{2}}\normalsize}\) substitution, we have

    \[{t = \tan \frac{x}{2},}\;\; \Rightarrow {x = 2\arctan t,\;\;}\kern0pt{dx = \frac{{2dt}}{{1 + {t^2}}},\;\;}\kern0pt{\sin x = \frac{{2t}}{{1 + {t^2}}},\;\;}\kern0pt{\cos x = \frac{{1 – {t^2}}}{{1 + {t^2}}}.}\]

    Then the integral in \(t-\)terms is written as

    \[{\int {\frac{{dx}}{{\sin x + \cos x + 1}}} }={ \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{\frac{{2t}}{{1 + {t^2}}} + \frac{{1 – {t^2}}}{{1 + {t^2}}} + 1}}} }={ \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{\frac{{2t + 1 – {t^2} + 1 + {t^2}}}{{1 + {t^2}}}}}} }={ \int {\frac{{2dt}}{{2t + 2}}} }={ \int {\frac{{dt}}{{t + 1}}} }={ \ln \left| {t + 1} \right| + C }={ \ln \left| {\tan \frac{x}{2} + 1} \right| + C.}\]

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    Problems 1-7
    Page 2
    Problems 8-18