I. Integration Using Derivatives
Note that: \(\frac{d}{{dx}}f\left( x \right) = f'\left( x \right)\). Thus,
\(\int {\frac{d}{{dx}}f\left( x \right)} = \int {f'\left( x \right)} \)
\(f\left( x \right) = \int {f'\left( x \right)} \)
Example: Find the integral value of \(y' = {\sec ^2}x\).
From derivatives
\(\frac{d}{{dx}}\tan x = {\sec ^2}x\)
Integrating both sides
\(\int {d\left( {\tan x} \right)} = \int {{{\sec }^2}xdx} \)
\(\tan x = \int {{{\sec }^2}xdx} \)
Thus:
\(y' = {\sec ^2}x\)
\(\int {d\left( y \right)} = \int {{{\sec }^2}x} dx\)
\(\ {y} = \tan x + c\)
2. Integration by trigonometric substitution
This method is applicable to problems involving the following forms:
1. \(\sqrt {{a^2} - {u^2}} \)
Assume or let u = a sin \(\theta \)
2. \(\sqrt {{a^2} + {u^2}} \)
Let u = a tan \(\theta \)
3. \(\sqrt {{u^2} - {a^2}} \)
Let u = a sec \(\theta \)
Example:
1. Integrate \(\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} \)
Show Solutions\(\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} \)
Let
\(x = 3\tan \theta \)
\(dx = 3{\sec ^2}\theta \)
\(x = 9{\tan ^2}x\)
\(\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} = \int {\frac{{3{{\sec }^2}\theta d\theta }}{{\sqrt {9 + 9{{\tan }^2}\theta } }}} \)
\(\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} = \frac{3}{3}\int {\frac{{{{\sec }^2}\theta d\theta }}{{\sqrt {1 + {{\tan }^2}\theta } }}} \)
\(\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} = \int {\frac{{{{\sec }^2}\theta d\theta }}{{\sqrt {{{\sec }^2}\theta } }}} \)
\(\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} = \int {\frac{{{{\sec }^2}\theta d\theta }}{{\sec \theta }}} \)
\(\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} = \int {\sec \theta d\theta } \)
\(\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} = \int {\sec \theta \left( {\frac{{\sec \theta + \tan \theta }}{{\sec \theta + \tan \theta }}} \right)} d\theta \)
\(\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} = \int {\frac{{{{\sec }^2}\theta + \sec \theta \tan \theta }}{{\sec \theta + \tan \theta }}} d\theta \)
\(\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} = \ln \left( {\sec \theta + \tan \theta } \right) + C\)
\(\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} = \ln \left( {\frac{{1 + \sin \theta }}{{\cos \theta }}} \right) + C\)
But from the assumption \(x = 3\tan \theta \)
\(\sin \theta = \frac{x}{{\sqrt {{x^2} + 9} }}\)
\(\cos \theta = \frac{3}{{\sqrt {{x^2} + 9} }}\)
\(\frac{{1 + \sin \theta }}{{\cos \theta }} = \frac{{1 + \frac{x}{{\sqrt {{x^2} + 9} }}}}{{\frac{3}{{\sqrt {{x^2} + 9} }}}}\)
\(\frac{{1 + \sin \theta }}{{\cos \theta }} = \frac{{\frac{{\sqrt {{x^2} + 9} + x}}{{\sqrt {{x^2} + 9} }}}}{{\frac{3}{{\sqrt {{x^2} + 9} }}}}\)
\(\frac{{1 + \sin \theta }}{{\cos \theta }} = \frac{{\sqrt {{x^2} + 9} + x}}{3}\)
Thus
\(\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }} = \ln \left( {\frac{{\sqrt {{x^2} + 9} + x}}{3}} \right) + C} \)
2. Integrate \(\int {y\sqrt {16 - {y^2}} dy} \)
Show Solutions\(\int\ y\sqrt {16 - {y^2}} dy\)
Let y =4 sin\(\theta \)
\(dy = 4\cos \theta d\theta \)
\({y^2} = 16{\sin ^2}\theta \)
\(\int\ y\sqrt {16 - {y^2}} dy = \int {\left( {4\sin \theta } \right)} \sqrt {16 - 16{{\sin }^2}\theta } \left( {4\cos \theta d\theta } \right)\)
\(\int\ y\sqrt {16 - {y^2}} dy = 64\int {\sin \theta \cos \theta \sqrt {1 - {{\sin }^2}\theta } d\theta } \)
\(\int\ y\sqrt {16 - {y^2}} dy = 64\int {\sin \theta {{\cos }^2}\theta d\theta } \)
\(\int\ y\sqrt {16 - {y^2}} dy = 64\left( {\frac{{{{\cos }^3}\theta }}{3}} \right) + C\)
But from the assumption y =4 sin\(\theta \)
\(\sin \theta = \frac{y}{4}\) and using the a right triangle,\(\sin \theta = \frac{{opp}}{{hyp}}\)
\(\cos \theta = \frac{{\sqrt {16 - {y^2}} }}{4}\)
\({\cos ^3}\theta = {\left( {\frac{{\sqrt {16 - {y^2}} }}{4}} \right)^3}\)
\({\cos ^3}\theta = \frac{{{{\left( {16 - {y^2}} \right)}^{\frac{3}{2}}}}}{{64}}\)
thus,\(\int {y\sqrt {16 - {y^2}} dy = 64\left( {\frac{{{{\cos }^3}\theta }}{3}} \right) + C} \)
\(\int {y\sqrt {16 - {y^2}} dy} = 64\left( {\frac{{{{\left( {16 - {y^2}} \right)}^{\frac{3}{2}}}}}{{64(3)}}} \right) + C\)
\(\int {y\sqrt {16 - {y^2}} } dy = \frac{{{{\left( {16 - {y^2}} \right)}^{\frac{3}{2}}}}}{3} + C\)
3. Integrate \(\int {\frac{{\sqrt {9{x^2} - 4} }}{x}} dx\)
Show Solution\(\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} \)
Let
\(3x = 2\sec \theta \)
\(9{x^2} = 4{\sec ^2}\theta \)
\(x = \frac{2}{3}\sec \theta \)
\(dx = \frac{2}{3}\sec \theta \tan \theta d\theta \)
\(\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = \int {\frac{{\sqrt {4{{\sec }^2}\theta - 4} }}{{\frac{2}{3}\sec \theta }}\left( {\frac{2}{3}\sec \theta \tan \theta d\theta } \right)} \)
\(\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = 2\int {\sqrt {{{\sec }^2}\theta - 1} \left( {\tan \theta d\theta } \right)} \)
\(\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = 2\int {\left( {{{\tan }^2}\theta d\theta } \right)} \)
\(\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = 2\int {\left( {{{\sec }^2}\theta - 1} \right)} d\theta \)
\(\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = 2\tan \theta - 2\theta + C\)
\(\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = 2\frac{{\sqrt {9{x^2} - 4} }}{2} - 2\arccos \left( {\frac{2}{{3x}}} \right) + C\)
\(\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = \sqrt {9{x^2} - 4} - 2\arccos \left( {\frac{2}{{3x}}} \right) + C\)
3. Integration by Parts
Integration by parts is from derivatives of a product.
\(\frac{d}{{dx}}\left( {uv} \right) = vdu + udv\)
\(\int {\frac{d}{{dx}}\left( {uv} \right)} = \int {vdu} + \int {udv} \)
\(uv = \int {vdu} + \int {udv} \)
\(\int {udv = uv - \int {vdu} } \)
How to integrate using integration by parts.
1. Assume for the values of u and dv, Sometimes this part is trial and error
2. Solve du and v
3. Substitute
Examples:
1. \(\int {x{e^x}dx} \)
Show SolutionLet u = x and \(dv = e^x dx\)
Thus, du = dx and \(v = e^x \)
\(\int {x{e^x}dx} \) = \(\int {udv} \)
\(\int {x{e^x}dx} \) = \(uv -\int{vdu} \)
\(\int {x{e^x}dx} \) = \(xe^x -\int{e^xdx} \)
\(\int {x{e^x}dx} \) = \(xe^x -e^x \) +c
2. \(\int {{e^x}\cos xdx} \)
Show SolutionLet \(u = cos x\) and \(e^xdx = dv\)
Thus\(du = -sin x dx\) and \(e^x = v\)
\(\int {{e^x}\cos xdx} \) = \(\int {udv}\)
\(\int {{e^x}\cos xdx} \) = \( uv -\int{vdu}\)
\(\int {{e^x}\cos xdx} \) = \(e^x cosx -\int{e^x(-sinx)dx}\)
\(\int {{e^x}\cos xdx} \) = \(e^x cosx +\int{e^x(sinx)dx}\)
Repeate the process of by parts on \(\int{e^x(sinx)dx}\)
Let u = sin x and \(e^xdx = dv\)
Thus du = cos x dx and \(e^x = v\)
\(\int{e^x(sinx)dx}\)=\(\int{udv}\)
\(\int{e^x(sinx)dx}\)=\(uv-\int{vdu}\)
\(\int{e^x(sinx)dx}\)=\(e^x sin x -\int{e^x cos xdx}\)
\(\int {{e^x}\cos xdx} \) = \(e^x cosx +\int{e^x(sinx)dx}\)
\(\int {{e^x}\cos xdx} \) = \(e^x cosx + e^x sin x -\int{e^x cos xdx}\)
\(2\int {{e^x}\cos xdx} \) = \(e^x cosx e^x sin x + C \)
\(\int {{e^x}\cos xdx} \) = \(\frac{e^x cosx e^x sin x + C}{2} \)