# 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

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

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Problems 1-2
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Problems 3-7