Trigonometric and Hyperbolic Substitutions
Solved Problems
Example 9.
Find the integral \[\int {\frac{{\sqrt {{x^2} - 16} }}{{{x^3}}} dx}.\]
Solution.
Using the trig substitution
\[x = 4\sec t,\;\; \Rightarrow dx = 4\tan t\sec tdt,\;\;\sqrt {{x^2} - 16} = 4\tan t,\;\;\cos t = \frac{4}{x},\;\;t = \arccos \frac{4}{x},\]
we rewrite the integral in the form
\[I = \int {\frac{{\sqrt {{x^2} - 16} }}{{{x^3}}}dx} = \int {\frac{{4\tan t \cdot 4\tan t\sec tdt}}{{{4^3}{{\sec }^3}t}}} = \frac{1}{4}\int {\frac{{{{\tan }^2}t\,dt}}{{{{\sec }^2}t}}} = \frac{1}{4}\int {{{\tan }^2}t\,{{\cos }^2}t\,dt} = \frac{1}{4}\int {{{\sin }^2}t\,dt} = \frac{1}{8}\int {\left( {1 - \cos 2t} \right)dt} .\]
Integrating yields:
\[I = \frac{1}{8}\int {\left( {1 - \cos 2t} \right)dt} = \frac{1}{8}\left( {t - \frac{{\sin 2t}}{2}} \right) + C = \frac{t}{8} - \frac{{\sin 2t}}{{16}} + C = \frac{t}{8} - \frac{{\sin t\cos t}}{8} + C = \frac{t}{8} - \frac{{\cos t\sqrt {1 - {{\cos }^2}t} }}{8} + C = \frac{1}{8}\arccos \frac{4}{x} - \frac{{\frac{4}{x}\sqrt {1 - {{\left( {\frac{4}{x}} \right)}^2}} }}{8} + C = \frac{1}{8}\arccos \frac{4}{x} - \frac{{\sqrt {{x^2} - 16} }}{{2{x^2}}} + C.\]
Example 10.
Evaluate the integral \[\int {\sqrt {2x - {x^2}} dx}.\]
Solution.
We can write the integral as
\[I = \int {\sqrt {2x - {x^2}} dx} = \int {\sqrt { - \left( {{x^2} - 2x} \right)} dx} = \int {\sqrt {1 - \left( {{x^2} - 2x + 1} \right)} dx} = \int {\sqrt {1 - {{\left( {x - 1} \right)}^2}} dx} .\]
Make the trigonometric substitution:
\[x - 1 = \sin t,\;\; \Rightarrow dx = \cos tdt,\;\; \Rightarrow t = \arcsin \left( {x - 1} \right).\]
Now we can calculate the integral:
\[I = \int {\sqrt {1 - {{\sin }^2}t} \cos tdt} = \int {{{\cos }^2}tdt} = \int {\frac{{1 + \cos 2t}}{2}dt} = \frac{t}{2} + \frac{1}{2}\int {\cos 2tdt} = \frac{t}{2} + \frac{{\sin 2t}}{4} + C = \frac{t}{2} + \frac{{\sin t\cos t}}{2} + C = \frac{t}{2} + \frac{{\sin t\sqrt {1 - {{\sin }^2}t} }}{2} + C = \frac{1}{2}\arcsin \left( {x - 1} \right) + \frac{{\left( {x - 1} \right)\sqrt {1 - {{\left( {x - 1} \right)}^2}} }}{2} + C = \frac{1}{2}\arcsin \left( {x - 1} \right) + \frac{1}{2}\left( {x - 1} \right)\sqrt {2x - {x^2}} + C.\]
Example 11.
Evaluate the integral \[\int {\sqrt {{x^2} + 4x + 3} dx}.\]
Solution.
We complete the square:
\[{x^2} + 4x + 3 = {x^2} + 4x + 4 - 1 = {\left( {x + 2} \right)^2} - 1\]
and make the substitution
\[u = x + 2,\;\; du = dx,\]
so the integral is written in the form
\[I = \int {\sqrt {{x^2} + 4x + 3} dx} = \int {\sqrt {{u^2} - 1} du} .\]
As we have an integral of type \(\int {R\left( {u,\sqrt {{u^2} - {r^2}} } \right)du},\) we can use the hyperbolic substitution
\[u = \cosh t,\;\; \Rightarrow du = \sinh tdt,\;\; t = \text{arccosh}\,u,\]
\[\sqrt {{u^2} - 1} = \sqrt {{{\cosh }^2}t - 1} = \sqrt {{{\sinh }^2}t} = \sinh t.\]
Then the integral becomes
\[I = \int {\sinh t\sinh tdt} = \int {{{\sinh }^2}tdt} .\]
Now we use the half-angle identity
\[{\sinh ^2}t = \frac{1}{2}\left( {\cosh 2t - 1} \right).\]
This yields:
\[I = \frac{1}{2}\int {\left( {\cosh 2t - 1} \right)dt} = \frac{1}{2}\left( {\frac{{\sinh 2t}}{2} - t} \right) + C = \frac{{\sinh 2t}}{4} - \frac{t}{2} + C = \frac{{\sinh t\cosh t}}{2} - \frac{t}{2} + C.\]
Returning to the original variable \(x,\) we obtain:
\[I = \frac{{\sinh t\cosh t}}{2} - \frac{t}{2} + C = \frac{{\sqrt {{{\cosh }^2}t - 1} \cosh t}}{2} - \frac{t}{2} + C = \frac{{u\sqrt {{u^2} - 1} }}{2} - \frac{{\text{arccosh}\,u}}{2} + C = \frac{{\left( {x + 2} \right)\sqrt {{{\left( {x + 2} \right)}^2} - 1} }}{2} - \frac{{\text{arccosh} \left( {x + 2} \right)}}{2} + C = \frac{{\left( {x + 2} \right)\sqrt {{x^2} + 4x + 3} }}{2} - \frac{{\text{arccosh} \left( {x + 2} \right)}}{2} + C.\]
Example 12.
Evaluate the integral \[\int {\sqrt {4 - {x^2}} dx}.\]
Solution.
Here we have an integral of type \(\int {R\left( {u,\sqrt {{r^2} - {u^2}} } \right)du},\) so we can make the trig substitution
\[x = 2\sin t,\;\; \Rightarrow dx = 2\cos tdt,\;\; t = \arcsin \frac{x}{2}.\]
The integral becomes
\[I = \int {\sqrt {4 - {x^2}} dx} = 2\int {\sqrt {4 - 4{{\sin }^2}t} \cos tdt} = 4\int {{{\cos }^2}tdt} .\]
We can rewrite it in the form
\[I = 4 \cdot \frac{1}{2}\int {\left( {1 + \cos 2t} \right)dt} = \int {\left( {2 + 2\cos 2t} \right)dt} = 2t + \frac{{\cancel{2}\sin 2t}}{\cancel{2}} + C = 2t + \sin 2t + C.\]
Express the sine function in terms of \(x:\)
\[\sin 2t = 2\sin \cos t = 2\sin t\sqrt {1 - {{\sin }^2}t} = 2 \cdot \frac{x}{2}\sqrt {1 - {{\left( {\frac{x}{2}} \right)}^2}} = x\sqrt {\frac{{4 - {x^2}}}{4}} = \frac{x}{2}\sqrt {4 - {x^2}} .\]
Hence, the integral is given by
\[I = 2t + \sin 2t + C = 2\arcsin \frac{x}{2} + \frac{x}{2}\sqrt {4 - {x^2}} + C.\]
Example 13.
Evaluate the integral \[\int {\sqrt {9{x^2} - 1} dx}.\]
Solution.
Here we make the hyperbolic substitution (for diversity):
\[3x = \cosh t,\;dx = {\frac{1}{3}} \sinh tdt.\]
As
\[1 + {\sinh ^2}t = {\cosh ^2}t,\]
the integral becomes
\[I = \int {\sqrt {9{x^2} - 1} dx} = \int {\left( {\sqrt {{{\cosh }^2}t - 1} \cdot \frac{1}{3}\sinh t} \right)dt} = \frac{1}{3}\int {{{\sinh }^2}tdt} .\]
Use the half-angle formula
\[{\sinh ^2}t = \frac{{\cosh 2t - 1}}{2}\]
to reduce the power of the integrand. Then
\[I = \frac{1}{3}\int {{{\sinh }^2}tdt} = \frac{1}{6}\int {\left( {\cosh 2t - 1} \right)dt} = \frac{1}{6} \cdot \frac{{\sinh 2t}}{2} - \frac{t}{6} + C = \frac{{\sinh t\cosh t}}{6} - \frac{t}{6} + C = \frac{{\sqrt {{\cosh^2}t - 1} \cosh t}}{6} - \frac{t}{6} + C = \frac{{\sqrt {9{x^2} - 1} \cdot 3x}}{6} - \frac{{\text{arccosh}\left( {3x} \right)}}{6} + C = \frac{{x\sqrt {9{x^2} - 1} }}{2} - \frac{{\text{arccosh}\left( {3x} \right)}}{6} + C.\]
Example 14.
Evaluate the integral \[\int {\sqrt {9 + {x^2}} dx}.\]
Solution.
We deal here with an integral of type \(\int {R\left( {u,\sqrt {{r^2} + {u^2}} } \right)du}.\) Let's try the hyperbolic substitution
\[x = 3\sinh t,\;\; \Rightarrow dx = 3\cosh tdt,\;\; \sqrt {9 + {x^2}} = 3\cosh t,\;\; t = \text{arcsinh} \frac{x}{3}.\]
Then the integral is written as
\[I = \int {\sqrt {9 + {x^2}} dx} = \int {3\cosh t \cdot 3\cosh tdt} = 9\int {{{\cosh }^2}tdt} .\]
Apply the hyperbolic identity
\[{\cosh ^2}t = \frac{1}{2}\left( {\cosh 2t + 1} \right).\]
Hence
\[I = \frac{9}{2}\int {\left( {\cosh 2t + 1} \right)dt} = \frac{9}{2}\left( {\frac{{\sinh 2t}}{2} + t} \right) + C = \frac{{9\sinh 2t}}{4} + \frac{{9t}}{2} + C = \frac{{9\sinh t\cosh t}}{2} + \frac{{9t}}{2} + C.\]
Returning back to the variable \(x,\) we have
\[I = \frac{{9\sinh t\cosh t}}{2} + \frac{{9t}}{2} + C = \frac{{9\sinh t\sqrt {1 + {{\sinh }^2}t} }}{2} + \frac{{9t}}{2} + C = \frac{{9 \cdot \frac{x}{3}\sqrt {1 + {{\left( {\frac{x}{3}} \right)}^2}} }}{2} + \frac{9}{2}\text{arcsinh} \frac{x}{3} + C = \frac{{x\sqrt {9 + {x^2}} }}{2} + \frac{9}{2}\text{arcsinh} \frac{x}{3} + C.\]
Example 15.
Evaluate the integral \[\int {\sqrt {{x^2} - 25} dx}.\]
Solution.
Here we can use the following substitution:
\[x = 5\cosh t,\;\; \Rightarrow dx = 5\sinh tdt,\;\; \sqrt {{x^2} - 25} = 5\sinh t,\;\; t = \text{arccosh} \frac{x}{5}.\]
Then the integral is written in the form
\[I = \int {\sqrt {{x^2} - 25} dx} = \int {5\sinh t \cdot 5\sinh tdt} = 25\int {{{\sinh }^2}tdt} .\]
We can easily integrate it using the identities
\[{\sinh ^2}t = \frac{1}{2}\left( {\cosh 2t - 1} \right),\]
\[\sinh 2t = 2\sinh t\cosh t,\]
\[{\cosh ^2}t - {\sinh ^2}t = 1.\]
Hence
\[I = 25\int {{{\sinh }^2}tdt} = \frac{{25}}{2}\int {\left( {\cosh 2t - 1} \right)dt} = \frac{{25}}{2}\left( {\frac{{\sinh 2t}}{2} - t} \right) + C = \frac{{25\sinh 2t}}{4} - \frac{{25t}}{2} + C = \frac{{25\sinh t\cosh t}}{2} - \frac{{25t}}{2} + C = \frac{{25\sqrt {{{\cosh }^2}t - 1} \cosh t}}{2} - \frac{{25t}}{2} + C = \frac{{25\sqrt {{{\left( {\frac{x}{5}} \right)}^2} - 1} \cdot \frac{x}{5}}}{2} - \frac{{25}}{2}\arcsin \frac{x}{5} + C = \frac{{x\sqrt {{x^2} - 25} }}{2} - \frac{{25}}{2}\arcsin \frac{x}{5} + C.\]
Example 16.
Compute the integral \[\int {x\sqrt {1 + {x^2}} dx}.\]
Solution.
Note that the best way to solve this integral is to use the direct substitution \(u = 1 + {x^2}.\) However, as this integral has the form \({R\left( {u,\sqrt {{r^2} + {u^2}} } \right)du},\) we can also evaluate it using the trigonometric substitution
\[x = \tan t,\;\; \Rightarrow dx = {\sec ^2}t\,dt,\;\; \sqrt {1 + {x^2}} = \sec t.\]
In this case, we obtain
\[I = \int {x\sqrt {1 + {x^2}} dx} = \int {\tan t\sec t\,{{\sec }^2}t\,dt} = \int {\tan t\,{{\sec }^3}t\,dt} = \int {\frac{{\sin tdt}}{{{{\cos }^4}t}}} = - \int {\frac{{d\left( {\cos t} \right)}}{{{{\cos }^4}t}}} = - \int {{{\left( {\cos t} \right)}^{ - 4}}d\left( {\cos t} \right)} = - \frac{{{{\left( {\cos t} \right)}^{ - 3}}}}{{\left( { - 3} \right)}} + C = \frac{1}{{3{{\cos }^3}t}} + C.\]
Recall the identity
\[{\cos ^2}t = \frac{1}{{1 + {{\tan }^2}t}}.\]
Then
\[I = \frac{1}{{3\,{{\cos }^3}t}} + C = \frac{1}{{\frac{3}{{\sqrt {{{\left( {1 + {{\tan }^2}t} \right)}^3}} }}}} + C = \frac{{\sqrt {{{\left( {1 + {{\tan }^2}t} \right)}^3}} }}{3} + C = \frac{{\sqrt {{{\left( {1 + {x^2}} \right)}^3}} }}{3} + C.\]
Example 17.
Evaluate the integral \[\int {\frac{{\sqrt {{x^2} + 1} }}{x} dx}.\]
Solution.
Using the trigonometric substitution
\[x = \tan t,\;dx = {\sec ^2}tdt,\]
we get
\[I = \int {\frac{{\sqrt {{x^2} + 1} }}{x}dx} = \int {\frac{{\sqrt {{{\tan }^2}t + 1} }}{{\tan t}}{{\sec }^2}tdt} = \int {\frac{{{{\sec }^3}t}}{{\tan t}}dt} = \int {\frac{{dt}}{{{\cos^2}t\sin t}}} = \int {\frac{{\sin tdt}}{{{\cos^2}t\,{{\sin }^2}t}}} = - \int {\frac{{d\left( {\cos t} \right)}}{{{\cos^2}t\left( {1 - {{\cos }^2}t} \right)}}} = \int {\frac{{d\left( {\cos t} \right)}}{{{\cos^2}t\left( {{{\cos }^2}t - 1} \right)}}} .\]
Now, we make the substitution \(\cos t = z.\) The integral becomes
\[I = \int {\frac{{dz}}{{{z^2}\left( {{z^2} - 1} \right)}}} .\]
Use the method of partial fractions to transform the integrand:
\[\frac{1}{{{z^2}\left( {{z^2} - 1} \right)}} = \frac{1}{{{z^2}\left( {z - 1} \right)\left( {z + 1} \right)}} = \frac{A}{{{z^2}}} + \frac{B}{z} + \frac{C}{{z - 1}} + \frac{D}{{z + 1}}.\]
Equate the coefficients:
\[1 = A\left( {{z^2} - 1} \right) + Bz\left( {{z^2} - 1} \right) + C{z^2}\left( {z + 1} \right) + D{z^2}\left( {z - 1} \right),\]
\[1 = A{z^2} - A + B{z^3} - Bz + C{z^3} + C{z^2} + D{z^3} - D{z^2},\]
\[1 = \left( {B + C + D} \right){z^3} + \left( {A + C + D} \right){z^2} - Bz - A.\]
Then
\[
\left\{ \begin{array}{l}
B + C + D = 0\\
A + C - D = 0\\
- B = 0\\
- A = 1
\end{array} \right.,\;\; \Rightarrow
\left\{ \begin{array}{l}
A = - 1\\
B = 0\\
C = \frac{1}{2}\\
D = - \frac{1}{2}
\end{array} \right..\]
Hence, we can write the integrand as
\[\frac{1}{{{z^2}\left( {{z^2} - 1} \right)}} = - \frac{1}{{{z^2}}} + \frac{1}{{2\left( {z - 1} \right)}} - \frac{1}{{2\left( {z + 1} \right)}}.\]
So, the initial integral is equal to
\[I = \int {\frac{{dz}}{{{z^2}\left( {{z^2} - 1} \right)}}} = - \int {\frac{{dz}}{{{z^2}}}} + \int {\frac{{dz}}{{2\left( {z - 1} \right)}}} - \int {\frac{{dz}}{{2\left( {z + 1} \right)}}} = \frac{1}{z} + \frac{1}{2}\ln \left| {z - 1} \right| - \frac{1}{2}\ln \left| {z + 1} \right| + C = \frac{1}{z} + \frac{1}{2}\ln \left| {\frac{{z - 1}}{{z + 1}}} \right| + C = \frac{1}{{\cos t}} + \frac{1}{2}\ln \left| {\frac{{\cos t - 1}}{{\cos t + 1}}} \right| + C.\]
Returning back to the variable \(x,\) we have
\[{\cos ^2}t = \frac{1}{{1 + {{\tan }^2}t}} = \frac{1}{{1 + {x^2}}},\]
The final answer is
\[I = \frac{1}{{\sqrt {\frac{1}{{1 + {x^2}}}} }} + \frac{1}{2}\ln \left| {\frac{{\sqrt {\frac{1}{{1 + {x^2}}}} - 1}}{{\sqrt {\frac{1}{{1 + {x^2}}}} + 1}}} \right| + C = \sqrt {1 + {x^2}} + \frac{1}{2}\ln \left| {\frac{{1 - \sqrt {1 + {x^2}} }}{{1 + \sqrt {1 + {x^2}} }}} \right| + C.\]
Example 18.
Evaluate the integral \[\int {\frac{{dx}}{{\sqrt {{x^2} + 2x} }}}.\]
Solution.
We rewrite this integral as follows:
\[I = \int {\frac{{dx}}{{\sqrt {{x^2} + 2x} }}} = \int {\frac{{dx}}{{\sqrt {{x^2} + 2x + 1 - 1} }}} = \int {\frac{{dx}}{{\sqrt {{{\left( {x + 1} \right)}^2} - 1} }}} .\]
Using the trig substitution
\[x + 1 = \sec t,\;\; \Rightarrow dx = \tan t\sec tdt,\;\; \sqrt {{{\left( {x + 1} \right)}^2} - 1} = \tan t,\;\; t = \arccos \frac{1}{{x + 1}},\]
we simplify the integral:
\[I = \int {\frac{{dx}}{{\sqrt {{{\left( {x + 1} \right)}^2} - 1} }}} = \int {\frac{{\cancel{\tan t}\sec tdt}}{{\cancel{\tan t}}}} = \int {\sec tdt} .\]
It is known that
\[\int {\sec tdt} = \ln \left| {\tan \left( {\frac{t}{2} + \frac{\pi }{4}} \right)} \right| + C.\]
Now we express the answer in terms of \(x.\) Since
\[\tan \left( {\frac{t}{2} + \frac{\pi }{4}} \right) = \frac{{\tan \frac{t}{2} + \tan \frac{\pi }{4}}}{{1 - \tan \frac{t}{2}\tan \frac{\pi }{4}}} = \frac{{\tan \frac{t}{2} + 1}}{{1 - \tan \frac{t}{2}}} = \frac{{\frac{{\sin \frac{t}{2}}}{{\cos \frac{t}{2}}} + 1}}{{1 - \frac{{\sin \frac{t}{2}}}{{\cos \frac{t}{2}}}}} = \frac{{\sin \frac{t}{2} + \cos \frac{t}{2}}}{{\cos \frac{t}{2} - \sin \frac{t}{2}}} = \frac{{{{\left( {\sin \frac{t}{2} + \cos \frac{t}{2}} \right)}^2}}}{{{{\cos }^2}\frac{t}{2} - {{\sin }^2}\frac{t}{2}}} = \frac{{1 + \sin t}}{{\cos t}} = \frac{{1 + \sqrt {1 - {{\cos }^2}t} }}{{\cos t}} = \frac{{1 + \sqrt {1 - {{\left( {\frac{1}{{x + 1}}} \right)}^2}} }}{{\frac{1}{{x + 1}}}} = \frac{{1 + \frac{{\sqrt {{{\left( {x + 1} \right)}^2} - 1} }}{{x + 1}}}}{{\frac{1}{{x + 1}}}} = x + 1 + \sqrt {{x^2} + 2x + \cancel{1} - \cancel{1}} = x + 1 + \sqrt {{x^2} + 2x} ,\]
the integral is given by
\[I = \ln \left| {x + 1 + \sqrt {{x^2} + 2x} } \right| + C.\]
