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

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

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

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

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

The integrals of the form

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

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

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

Trigonometric substitution:

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

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

Trigonometric substitution:

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

Hyperbolic substitution:

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

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

Trigonometric substitution:

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

Hyperbolic substitution:

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

Remarks.

Solved Problems

Click or tap a problem to see the solution.

Example 1

Find the integral \[\int {\frac{{dx}}{{{x^2}\sqrt {1 - {x^2}} }}}.\]

Example 2

Evaluate the integral \[\int {\frac{{\sqrt {{a^2} - {x^2}} dx}}{{{x^2}}}}.\]

Example 3

Compute the integral \[\int {\frac{{xdx}}{{\sqrt {5 - {x^2}} }}}.\]

Example 4

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

Example 5

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

Example 6

Calculate the integral \[\int {\frac{{dx}}{{\sqrt {{{\left( {{x^2} - 8} \right)}^3}} }}}.\]

Example 7

Calculate the integral \[\int {\frac{{dx}}{{\sqrt {\left( {x - a} \right)\left( {b - x} \right)} }}}.\]

Example 8

Evaluate the integral \[\int {\frac{{\sqrt {{x^2} - {a^2}} }}{x} dx}.\]

Example 1.

Find the integral \[\int {\frac{{dx}}{{{x^2}\sqrt {1 - {x^2}} }}}.\]

Solution.

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

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

The integral can be easily evaluated:

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

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 \[\int {\frac{{\sqrt {{a^2} - {x^2}} dx}}{{{x^2}}}}.\]

Solution.

We make the substitution:

\[x = a\sin t,\;\; dx = a\cos tdt,\;\; t = \arcsin \frac{x}{a}.\]

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 \[\int {\frac{{xdx}}{{\sqrt {5 - {x^2}} }}}.\]

Solution.

We make the following trig substitution:

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

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 \[\int {\frac{{dx}}{{\sqrt {{{\left( {{x^2} + 1} \right)}^3}} }}}.\]

Solution.

We make the substitution

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

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 \[\int {\frac{{dx}}{{\sqrt {{{\left( {{a^2} + {x^2}} \right)}^3}} }}}.\]

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 \[\int {\frac{{dx}}{{\sqrt {{{\left( {{x^2} - 8} \right)}^3}} }}}.\]

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

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

Example 7.

Calculate the integral \[\int {\frac{{dx}}{{\sqrt {\left( {x - a} \right)\left( {b - x} \right)} }}}.\]

Solution.

We make the substitution:

\[x - a = \left( {b - a} \right){\sin ^2}t.\]

Hence,

\[x = a + \left( {b - a} \right){\sin ^2}t,\;\; dx = 2\left( {b - a} \right)\sin t\cos tdt,\]
\[ \Rightarrow b - x = b - a - \left( {b - a} \right){\sin ^2}t = \left( {b - a} \right)\left( {1 - {{\sin }^2}t} \right) = \left( {b - a} \right){\cos^2}t.\]

Then the integral becomes

\[I = \int {\frac{{dx}}{{\sqrt {\left( {x - a} \right)\left( {b - x} \right)} }}} = \int {\frac{{2\left( {b - a} \right)\sin t\cos tdt}}{{\sqrt {\left( {b - a} \right){{\sin }^2}t\left( {b - a} \right){{\cos }^2}t} }}} = \int {\frac{{2\left( \cancel{b - a} \right)\cancel{\sin t}\cancel{\cos t}dt}}{{\left( \cancel{b - a} \right)\cancel{\sin t}\cancel{\cos t}}}} = 2\int {dt} = 2t + C.\]

Returning back to the variable \(x:\)

\[{\sin ^2}t = \frac{{x - a}}{{b - a}},\;\; \sin t = \sqrt {\frac{{x - a}}{{b - a}}} ,\;\; t = \arcsin \sqrt {\frac{{x - a}}{{b - a}}} ,\]

we obtain the complete answer:

\[I = 2\arcsin \sqrt {\frac{{x - a}}{{b - a}}} + C.\]

Example 8.

Evaluate the integral \[\int {\frac{{\sqrt {{x^2} - {a^2}} }}{x} dx}.\]

Solution.

We make the trigonometric substitution:

\[x = a\sec t,\;dx = a\tan t\sec tdt.\]

Calculate the integral using the identity

\[1 + {\tan ^2}t = {\sec ^2}t:\]
\[I = \int {\frac{{\sqrt {{x^2} - {a^2}} }}{x}dx} = \int {\frac{{\sqrt {{a^2}{{\sec }^2}t - {a^2}} }}{{a\sec t}} }\cdot { a\tan t\sec tdt} = \int {\frac{{a\tan t}}{{a\sec t}}a\tan t\sec tdt} = a\int {{{\tan }^2}tdt} = a\int {\left( {{\sec^2}t - 1} \right)dt} = a\tan t - at + C = a\sqrt {{{\sec }^2}t - 1} - at + C.\]

For \(t,\) we have the following expression:

\[x = \frac{a}{{\cos t}},\;\; \cos t = \frac{a}{x},\;\; t = \arccos \frac{a}{x},\]

Hence, returning back to the variable \(x,\) we obtain:

\[I = a\sqrt {{{\sec }^2}t - 1} - at + C = a\sqrt {\frac{{{x^2}}}{{{a^2}}} - 1} - a\arccos \frac{a}{x} + C = \sqrt {{x^2} - {a^2}} - a\arccos \frac{a}{x} + C.\]

See more problems on Page 2.

Page 1 Page 2