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 = b² − 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
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.
