Calculus

Integration of Functions

Trigonometric and Hyperbolic Substitutions

Page 1
Problems 1-2
Page 2
Problems 3-9

In this section we consider integrals of the form \({\large\int\normalsize} {R\left( {x,\sqrt {a{x^2} + bx + c} } \right)dx} ,\) where \(R\) is a rational function of \(x\) and the radical \({\sqrt {a{x^2} + bx + c} }.\)

To calculate such an integral, we need first to complete the square in the quadratic expression:

\[
{a{x^2} + bx + c }
= {a\left( {{{\left( {x + \frac{b}{{2a}}} \right)}^2} }\right.}+{\left.{ \left( {c – \frac{{{b^2}}}{{4a}}} \right)} \right).}
\]

Making the substitution \(u = x + \large\frac{b}{{2a}}\normalsize,\) \(du = dx,\) we can obtain one of the three possible integrals, depending on the values of the coefficients \(a, b\) and \(c:\)

  1. \({\large\int\normalsize} {R\left( {u,\sqrt {{r^2} – {u^2}} } \right)du} \)
  2. \({\large\int\normalsize} {R\left( {u,\sqrt {{r^2} + {u^2}} } \right)du} \)
  3. \({\large\int\normalsize} {R\left( {u,\sqrt {{u^2} – {r^2}} } \right)du} \)

Then we use the following trigonometric or hyperbolic substitutions to simplify the integrals:

1. Integrals of the form \({\large\int\normalsize} {R\left( {u,\sqrt {{r^2} – {u^2}} } \right)du} \)

Trigonometric substitution:

\[
{u = r\sin t,\;\;}\kern-0.3pt
{du = r\cos tdt,\;\;}\kern-0.3pt
{\sqrt {{r^2} – {u^2}} = r\cos t,\;\;}\kern-0.3pt
{t = \arcsin \left( {\frac{u}{r}} \right).}
\]

2. Integrals of the form \({\large\int\normalsize} {R\left( {u,\sqrt {{r^2} + {u^2}} } \right)du} \)

Trigonometric substitution:

\[
{u = r\tan t,\;\;}\kern-0.3pt
{du = r\,{\sec ^2}tdt,\;\;}\kern-0.3pt
{\sqrt {{r^2} + {u^2}} = r\sec t,\;\;}\kern-0.3pt
{t = \arctan \left( {\frac{u}{r}} \right).}
\]

Hyperbolic substitution:

\[
{u = r\sinh t,\;\;}\kern-0.3pt
{du = r\cosh tdt,\;\;}\kern-0.3pt
{\sqrt {{r^2} + {u^2}} = r\cosh t,\;\;}\kern-0.3pt
{t = \text{arcsinh} \left( {\frac{u}{r}} \right).}
\]

3. Integrals of the form \({\large\int\normalsize} {R\left( {u,\sqrt {{u^2} – {r^2}} } \right)du} \)

Trigonometric substitution:

\[
{u = r\sec t,\;\;}\kern-0.3pt
{du = r\tan t\sec tdt,\;\;}\kern-0.3pt
{\sqrt {{u^2} – {r^2}} = r\tan t,\;\;}\kern-0.3pt
{t = \arccos \left( {\frac{r}{u}} \right).}
\]

Hyperbolic substitution:

\[
{u = r\cosh t,\;\;}\kern-0.3pt
{du = r\sinh tdt,\;\;}\kern-0.3pt
{\sqrt {{u^2} – {r^2}} = r\sinh t,\;\;}\kern-0.3pt
{t = \text{arccosh} \left( {\frac{u}{r}} \right).}
\]

Remarks:

  • Instead of the trigonometric substitutions in cases \(1, 2, 3\) you can use the substitutions
    \(x = r\cos t,\) \(x = r\cot t,\) \(x = r\csc t,\) respectively.
  • Using the formulas given above, we consider only positive values of the root. For example, in strict writing
    \[
    {\sqrt {{r^2} – {u^2}} }
    = {\sqrt {{r^2} – {r^2}{{\cos }^2}t} }
    = {\sqrt {{r^2}{\sin^2}t} }
    = {\left| {r\sin t} \right|.}
    \]

    We suppose here that \(\left| {r\sin t} \right| \) \(= r\sin t.\)

Solved Problems

Click on problem description to see solution.

 Example 1

Evaluate the integral \({\large\int\normalsize} {\large\frac{{\sqrt {{a^2} – {x^2}} dx}}{{{x^2}}}\normalsize} .\)

 Example 2

Evaluate the integral \({\large\int\normalsize} {\large\frac{{dx}}{{\sqrt {{{\left( {{a^2} + {x^2}} \right)}^3}} }}\normalsize} .\)

 Example 3

Calculate the integral \({\large\int\normalsize} {\large\frac{{dx}}{{\sqrt {{{\left( {{x^2} – 8} \right)}^3}} }}\normalsize} .\)

 Example 4

Calculate the integral \({\large\int\normalsize} {\large\frac{{dx}}{{\sqrt {\left( {x – a} \right)\left( {b – x} \right)} }}\normalsize} .\)

 Example 5

Evaluate the integral \({\large\int\normalsize} {\large\frac{{\sqrt {{x^2} – {a^2}} }}{x}\normalsize dx} .\)

 Example 6

Evaluate the integral \({\large\int\normalsize} {\sqrt {2x – {x^2}} dx} .\)

 Example 7

Evaluate the integral \({\large\int\normalsize} {\sqrt {9{x^2} – 1} dx} .\)

 Example 8

Calculate the integral \({\large\int\normalsize} {\sqrt {{x^2} + 2x + 3}\,dx} .\)

 Example 9

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

Example 1.

Evaluate the integral \({\large\int\normalsize} {\large\frac{{\sqrt {{a^2} – {x^2}} dx}}{{{x^2}}}\normalsize} .\)

Solution.

We make the substitution:

\[
{x = a\sin t,\;\;}\kern-0.3pt
{dx = a\cos tdt,\;\;}\kern0pt
{t = \arcsin \frac{x}{a}.}
\]

Then

\[
{\int {\frac{{\sqrt {{a^2} – {x^2}} dx}}{{{x^2}}}} }
= {\int {\frac{{\sqrt {{a^2} – {a^2}{{\sin }^2}t} }}{{{a^2}{{\sin }^2}t}}a\cos tdt} }
= {\int {\frac{{a\cos t}}{{{a^2}{{\sin }^2}t}}a\cos tdt} }
= {\int {{{\cot }^2}tdt} = \int {\left( {{{\csc }^2}t – 1} \right)dt} }
= { – \cot t – t + C }={{ – \frac{{\sqrt {1 – {{\sin }^2}t} }}{{\sin t}} – t }+{ C }}
= {{ – \frac{{\sqrt {1 – \frac{{{x^2}}}{{{a^2}}}} }}{{\frac{x}{a}}} – \arcsin \frac{x}{a} }+{ C }}
= {{ – \frac{{\sqrt {{a^2} – {x^2}} }}{x} – \arcsin \frac{x}{a} }+{ C.}}
\]

To simplify the integral, we used here the trigonometric formula \({\cot ^2}t = {\csc ^2}t – 1.\)

Example 2.

Evaluate the integral \({\large\int\normalsize} {\large\frac{{dx}}{{\sqrt {{{\left( {{a^2} + {x^2}} \right)}^3}} }}\normalsize} .\)

Solution.

We make the hyperbolic substitution: \(x = a\sinh t,\) \(dx = a\cosh tdt.\) Using the hyperbolic identity \(1 + {\sinh ^2}t \) \(= {\cosh ^2}t,\) we can write:

\[
{\int {\frac{{dx}}{{\sqrt {{{\left( {{a^2} + {x^2}} \right)}^3}} }}} }
= {\int {\frac{{a\cosh tdt}}{{\sqrt {{{\left( {{a^2} + {a^2}{{\sinh }^2}t} \right)}^3}} }}} }
= {\int {\frac{{a\cosh tdt}}{{{a^3}{{\cosh }^3}t}}} }
= {\frac{1}{{{a^2}}}\int {{{\text{sech}}^2}tdt} }
= {\frac{1}{{{a^2}}}\tanh t + C }
= {\frac{1}{{{a^2}}}\frac{{\sinh t}}{{\cosh t}} + C }
= {\frac{1}{{{a^2}}}\frac{{\sinh t}}{{\sqrt {1 + {\sinh^2}t} }} + C }
= {\frac{1}{{{a^2}}}\frac{{\frac{x}{a}}}{{\sqrt {1 + \frac{{{x^2}}}{{{a^2}}}} }} + C }
= {\frac{1}{{{a^2}}}\frac{x}{{\sqrt {{a^2} + {x^2}} }} + C.}
\]
Page 1
Problems 1-2
Page 2
Problems 3-9