The Indefinite Integral and Basic Rules of Integration
Solved Problems
Example 9.
Calculate \[\int {\frac{{4dx}}{{2 + 3{x^2}}}}.\]
Solution.
We use the table integral
\[\int {\frac{{dx}}{{{a^2} + {x^2}}} = \frac{1}{a}} \arctan {\frac{x}{a}} + C.\]
Then
\[\int {\frac{{4dx}}{{2 + 3{x^2}}}} = 4\int {\frac{{dx}}{{3\left( {\frac{2}{3} + {x^2}} \right)}}} = \frac{4}{3}\int {\frac{{dx}}{{{{\left( {\sqrt {\frac{2}{3}} } \right)}^2} + {x^2}}}} = \frac{4}{3} \cdot \frac{1}{{\sqrt {\frac{2}{3}} }}\arctan \frac{x}{{\sqrt {\frac{2}{3}} }} + C = \frac{4}{{\sqrt 6 }}\arctan \frac{{\sqrt 3 x}}{{\sqrt 2 }} + C.\]
Example 10.
Find the indefinite integral \[\int {\frac{{{x^2}}}{{1 + {x^2}}} dx}.\]
Solution.
We write the integral as the sum of two integrals:
\[I = \int {\frac{{{x^2}}}{{1 + {x^2}}}dx} = \int {\frac{{1 + {x^2} - 1}}{{1 + {x^2}}}dx} = \int {\left( {\frac{{1 + {x^2}}}{{1 + {x^2}}} - \frac{1}{{1 + {x^2}}}} \right)dx} = \int {\left( {1 - \frac{1}{{1 + {x^2}}}} \right)dx} = \int {dx} - \int {\frac{{dx}}{{1 + {x^2}}}} .\]
Hence
\[I = \int {dx} - \int {\frac{{dx}}{{1 + {x^2}}}} = x - \arctan x + C.\]
Example 11.
Evaluate the integral \[\int {\frac{{dx}}{{1 + 2{x^2}}}}.\]
Solution.
We can write the integral as follows:
\[I = \int {\frac{{dx}}{{1 + 2{x^2}}}} = \int {\frac{{dx}}{{2\left( {\frac{1}{2} + {x^2}} \right)}}} = \frac{1}{2}\int {\frac{{dx}}{{\frac{1}{2} + {x^2}}}} = \frac{1}{2}\int {\frac{{dx}}{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} + {x^2}}}} .\]
Using the table integral
\[\int {\frac{{dx}}{{{a^2} + {x^2}}}} = {\frac{1}{a}}\arctan {\frac{x}{a}},\]
we have
\[I = \frac{1}{2}\int {\frac{{dx}}{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} + {x^2}}}} = \frac{1}{2} \cdot \frac{1}{{\frac{1}{{\sqrt 2 }}}}\arctan \frac{x}{{\frac{1}{{\sqrt 2 }}}} + C = \frac{{\sqrt 2 }}{2}\arctan \left( {\sqrt 2 x} \right) + C = \frac{1}{{\sqrt 2 }}\arctan \left( {\sqrt 2 x} \right) + C.\]
Example 12.
Find the integral \[\int {\frac{{\pi dx}}{{\sqrt {\pi - {x^2}} }}}.\]
Solution.
Using the table integral
\[\int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }}} = \arcsin {\frac{x}{a}} + C,\]
we obtain
\[\int {\frac{{\pi dx}}{{\sqrt {\pi - {x^2}} }}} = \pi \int {\frac{{dx}}{{\sqrt {{{\left( {\sqrt \pi } \right)}^2} - {x^2}} }}} = \pi \arcsin \frac{x}{{\sqrt \pi }} + C.\]
Example 13.
Calculate the integral \[\int {\left( {2\cos x - 5\sin x} \right)dx}.\]
Solution.
Using the sum rule, the constant multiple rule and the table of basic integrals, we have
\[\int {\left( {2\cos x - 5\sin x} \right)dx} = \int {2\cos xdx} - \int {5\sin xdx} = 2\int {\cos xdx} - 5\int {\sin xdx} = 2 \cdot \sin x - 5 \cdot \left( { - \cos x} \right) + C = 2\sin x + 5\cos x + C.\]
Example 14.
Evaluate the integral \[\int {\frac{{dx}}{{\sqrt {1 - \frac{{{x^2}}}{2}} }}}.\]
Solution.
We use some algebra to reduce the integral to the standard form:
\[I = \int {\frac{{dx}}{{\sqrt {1 - \frac{{{x^2}}}{2}} }}} = \int {\frac{{dx}}{{\sqrt {\frac{1}{2}\left( {2 - {x^2}} \right)} }}} = \int {\frac{{dx}}{{\sqrt {\frac{1}{2}} \sqrt {2 - {x^2}} }}} = \sqrt 2 \int {\frac{{dx}}{{\sqrt {2 - {x^2}} }}} = \sqrt 2 \int {\frac{{dx}}{{\sqrt {{{\left( {\sqrt 2 } \right)}^2} - {x^2}} }}} .\]
The expression above contains the table integral
\[\int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }}} = \arcsin {\frac{x}{a}},\]
so we have
\[I = \sqrt 2 \int {\frac{{dx}}{{\sqrt {{{\left( {\sqrt 2 } \right)}^2} - {x^2}} }}} = \sqrt 2 \arcsin \frac{x}{{\sqrt 2 }} + C.\]
Example 15.
Calculate the integral \[\int {{{\tan }^2}xdx}.\]
Solution.
Since
\[{\tan ^2}x = {\sec ^2}x - 1,\]
the integral is
\[\int {{{\tan }^2}xdx} = \int {\left( {{{\sec }^2}x - 1} \right)dx} = \int {{{\sec }^2}xdx} - \int {dx} = \tan x - x + C.\]
Example 16.
Calculate the integral \[\int {{{\cot }^2}xdx}.\]
Solution.
We use the trig identity
\[\frac{1}{{{{\sin }^2}x}} - {\cot ^2}x = 1,\;\; \Rightarrow {\cot ^2}x = \frac{1}{{{{\sin }^2}x}} - 1.\]
Then the integral is written as the sum of two integrals:
\[I = \int {{{\cot }^2}xdx} = \int {\left( {\frac{1}{{{{\sin }^2}x}} - 1} \right)dx} = \int {\frac{{dx}}{{{{\sin }^2}x}}} - \int {dx} .\]
Hence
\[I = \int {\frac{{dx}}{{{{\sin }^2}x}}} - \int {dx} = - \cot x - x + C.\]
Example 17.
Find the integral \[\int {\frac{{dx}}{{{\sin^2}2x}}}\] without using a substitution.
Solution.
Using the double angle formula
\[\sin 2x = 2\sin x\cos x\]
and Pythagorean identity
\[{\sin^2}x + {\cos ^2}x = 1,\]
we can write:
\[\int {\frac{{dx}}{{{\sin^2}2x}}} = \frac{1}{4}\int {\frac{{dx}}{{{\sin^2}x{{\cos }^2}x}}} = \frac{1}{4}\int {\frac{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)dx}}{{{\sin^2}x{{\cos }^2}x}}} = \frac{1}{4}\int {\left( {\frac{1}{{{{\cos }^2}x}} + \frac{1}{{{\sin^2}x}}} \right)dx} = \frac{1}{4}\int {{{\sec }^2}xdx} + \frac{1}{4}\int {{\csc^2}xdx} = \frac{1}{4}\tan x - \frac{1}{4}\cot x + C = \frac{1}{4}\left( {\tan x - \cot x} \right)+ C.\]
