Calculus

Integration of Functions

Integration of Functions Logo

Trigonometric and Hyperbolic Substitutions

In this section we consider the integration of functions containing a radical of the form \(\sqrt {a{x^2} + bx + c}.\)

When calculating such an integral, we first need to complete the square in the quadratic expression:

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

where \(D = {b^2} – 4ac.\)

Making the substitution

\[{u = x + \frac{b}{{2a}},\;\;}\kern0pt{du = dx,}\]

we can obtain one of the following three expressions depending on the signs of \(a\) and \(D:\)

\[{\sqrt {{r^2} – {u^2}} ,\;\;}\kern0pt{\sqrt {{r^2} + {u^2}} ,\;\;}\kern0pt{\sqrt {{u^2} – {r^2}} ,}\]

where \(r = \sqrt {\left| {\large{\frac{D}{{4{a^2}}}}\normalsize} \right|} \gt 0.\)

The integrals of the form

\[{\int {R\left( {u,\sqrt {{r^2} – {u^2}} } \right)du} ,\;\;}\kern0pt{\int {R\left( {u,\sqrt {{r^2} + {u^2}} } \right)du} ,\;\;}\kern0pt{\int {R\left( {u,\sqrt {{u^2} – {r^2}} } \right)du} ,}\]

where \(R\) denotes a rational function, can be evaluated using trigonometric or hyperbolic substitutions.

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 the positive square roots. 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 or tap a problem to see the solution.

Example 1

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

Example 2

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

Example 3

Compute the integral \(\int {\large{\frac{{xdx}}{{\sqrt {5 – {x^2}} }}}\normalsize}.\)

Example 4

Find the integral \(\int {\large{\frac{{dx}}{{\sqrt {{{\left( {{x^2} + 1} \right)}^3}} }}}\normalsize}.\)

Example 5

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

Example 6

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

Example 7

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

Example 8

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

Example 9

Find the integral \(\int {\large{\frac{{\sqrt {{x^2} – 16} }}{{{x^3}}}}\normalsize dx}.\)

Example 10

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

Example 11

Evaluate the integral \(\int {\sqrt {{x^2} + 4x + 3} dx}.\)

Example 12

Evaluate the integral \(\int {\sqrt {4 – {x^2}} dx}.\)

Example 13

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

Example 14

Evaluate the integral \(\int {\sqrt {9 + {x^2}} dx}.\)

Example 15

Evaluate the integral \(\int {\sqrt {{x^2} – 25} dx}.\)

Example 16

Compute the integral \(\int {x\sqrt {1 + {x^2}} dx}.\)

Example 17

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

Example 18

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

Example 1.

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

Solution.

Let’s try the trig substitution \(x = \sin t.\) Then

\[{dx = \cos tdt,\;\;}\kern0pt{\sqrt {1 – {x^2}} }={ \sqrt {1 – {{\sin }^2}t} }={ \cos t.}\]

The integral can be easily evaluated:

\[\require{cancel}{I = \int {\frac{{dx}}{{{x^2}\sqrt {1 – {x^2}} }}} }={ \int {\frac{{\cos tdt}}{{{{\sin }^2}t\cos t}}} }={ \int {\frac{{dt}}{{{{\sin }^2}t}}} }={ – \cot t + C.}\]

Now we should express the result in terms of the original variable \(x:\)

\[{I = – \cot t + C }={ – \frac{{\cos t}}{{\sin t}} + C }={ – \frac{{\sqrt {1 – {{\sin }^2}t} }}{{\sin t}} + C }={ – \frac{{\sqrt {1 – {x^2}} }}{x} + C.}\]

Example 2.

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 identity \({\cot ^2}t = {\csc ^2}t – 1.\)

Example 3.

Compute the integral \(\int {\large{\frac{{xdx}}{{\sqrt {5 – {x^2}} }}}\normalsize}.\)

Solution.

We make the following trig substitution:

\[{x = \sqrt 5 \sin t,}\;\; \Rightarrow {dx = \sqrt 5 \cos tdt,\;\;}\kern0pt{\sqrt {5 – {x^2}} = \sqrt 5 \cos t.}\]

Then the integral becomes:

\[{I = \int {\frac{{xdx}}{{\sqrt {5 – {x^2}} }}} }={ \int {\frac{{\sqrt 5 \sin t \cdot \cancel{\sqrt 5} \cancel{\cos t}dt}}{{\cancel{\sqrt 5} \cancel{\cos t}}}} }={ \sqrt 5 \int {\sin tdt} }={ – \sqrt 5 \cos t + C.}\]

Express the answer in terms of \(x:\)

\[{I = – \sqrt 5 \cos t + C }={ – \sqrt 5 \sqrt {1 – {{\sin }^2}t} + C }={ – \sqrt 5 \sqrt {1 – {{\left( {\frac{x}{{\sqrt 5 }}} \right)}^2}} + C }={ – \sqrt {5 – {x^2}} + C.}\]

Example 4.

Find the integral \(\int {\large{\frac{{dx}}{{\sqrt {{{\left( {{x^2} + 1} \right)}^3}} }}}\normalsize}.\)

Solution.

We make the substitution

\[{x = \tan t,}\;\; \Rightarrow {dx = {\sec ^2}tdt,\;\;}\kern0pt{\sqrt {{{\left( {{x^2} + 1} \right)}^3}} }={ {\left( {\sqrt {{x^2} + 1} } \right)^3} }={ {\left( {\sqrt {{{\tan }^2}t + 1} } \right)^3} }={ {\sec ^3}t.}\]

This yields:

\[{I = \int {\frac{{dx}}{{\sqrt {{{\left( {{x^2} + 1} \right)}^3}} }}} }={ \int {\frac{{{{\sec }^2}tdt}}{{{{\sec }^3}t}}} }={ \int {\frac{{dt}}{{\sec t}}} }={ \int {\cos tdt} }={ \sin t + C.}\]

Returning to the variable \(x,\) we get:

\[{I = \sin t + C }={ \frac{{\tan t}}{{\sqrt {{{\tan }^2}t + 1} }} + C }={ \frac{x}{{\sqrt {{x^2} + 1} }} + C.}\]

Example 5.

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:

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

Example 6.

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

Solution.

To find the integral, we make the substitution: \(x = \sqrt 8 \sec t,\) \(dx = \sqrt 8 \tan t\sec tdt.\) Using the identity \({\sec ^2}t – 1 \) \(= {\tan ^2}t,\) we have

\[
{I = \int {\frac{{dx}}{{\sqrt {{{\left( {{x^2} – 8} \right)}^3}} }}} }
= {\int {\frac{{\sqrt 8 \tan t\sec tdt}}{{\sqrt {{{\left( {8{{\sec }^2}t – 8} \right)}^3}} }}} }
= {\frac{{\sqrt 8 }}{{\sqrt {{8^3}} }}\int {\frac{{\tan t\sec tdt}}{{{{\tan }^3}t}}} }
= {\frac{1}{8}\int {\frac{{\cos tdt}}{{{{\sin }^2}t}}} }
= {\frac{1}{8}\int {\frac{{d\left( {\sin t} \right)}}{{{{\sin }^2}t}}} }
= { – \frac{1}{{8\sin t}} + C.}
\]

Express \(\sin t\) in terms of \(x:\)

\[
{\sin t = \sqrt {1 – {{\cos }^2}t} }
= {\sqrt {1 – \frac{1}{{{{\sec }^2}t}}} }
= {\frac{{\sqrt {{{\sec }^2}t – 1} }}{{\sec t}} }
= {\frac{{\sqrt {\frac{{{x^2}}}{8} – 1} }}{{\frac{x}{{\sqrt 8 }}}} }
= {\frac{{\sqrt {{x^2} – 8} }}{x}.}
\]

Hence, the integral is

\[
{I }={ – \frac{1}{{8\frac{{\sqrt {{x^2} – 8} }}{x}}} + C }
= { – \frac{x}{{8\sqrt {{x^2} – 8} }} }+{ C.}
\]

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Problems 1-6
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Problems 7-18