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  • Integral Calculus Solutions

    Topics || Problems

    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 \(\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} \)

    \(\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} \)

    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 \(\int {y\sqrt {16 - {y^2}} dy} \)

    \(\int\ y\sqrt {16 - {y^2}} dy\)

    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,\(\sin \theta = \frac{{opp}}{{hyp}}\)

    integral calculus

    \(\cos \theta = \frac{{\sqrt {16 - {y^2}} }}{4}\)

    \({\cos ^3}\theta = {\left( {\frac{{\sqrt {16 - {y^2}} }}{4}} \right)^3}\)

    \({\cos ^3}\theta = \frac{{{{\left( {16 - {y^2}} \right)}^{\frac{3}{2}}}}}{{64}}\)

    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 \(\int {\frac{{\sqrt {9{x^2} - 4} }}{x}} dx\)

    \(\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} \)

    Let

    \(3x = 2\sec \theta \)

    \(9{x^2} = 4{\sec ^2}\theta \)

    \(x = \frac{2}{3}\sec \theta \)

    \(dx = \frac{2}{3}\sec \theta \tan \theta d\theta \)

    \(\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = \int {\frac{{\sqrt {4{{\sec }^2}\theta - 4} }}{{\frac{2}{3}\sec \theta }}\left( {\frac{2}{3}\sec \theta \tan \theta d\theta } \right)} \)

    \(\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = 2\int {\sqrt {{{\sec }^2}\theta - 1} \left( {\tan \theta d\theta } \right)} \)

    \(\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = 2\int {\left( {{{\tan }^2}\theta d\theta } \right)} \)

    \(\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = 2\int {\left( {{{\sec }^2}\theta - 1} \right)} d\theta \)

    \(\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = 2\tan \theta - 2\theta + C\)

    integration by substitution

    \(\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = 2\frac{{\sqrt {9{x^2} - 4} }}{2} - 2\arccos \left( {\frac{2}{{3x}}} \right) + C\)

    \(\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = \sqrt {9{x^2} - 4} - 2\arccos \left( {\frac{2}{{3x}}} \right) + C\)

    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. \(\int {x{e^x}dx} \)

    Let 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. \(\int {{e^x}\cos xdx} \)

    Let \(u = cos x\) 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 \(e^x = v\)

    \(\int{e^x(sinx)dx}\)=\(\int{udv}\)

    \(\int{e^x(sinx)dx}\)=\(uv-\int{vdu}\)

    \(\int{e^x(sinx)dx}\)=\(e^x sin x -\int{e^x cos xdx}\)

    \(\int {{e^x}\cos xdx} \) = \(e^x cosx +\int{e^x(sinx)dx}\)

    \(\int {{e^x}\cos xdx} \) = \(e^x cosx + e^x sin x -\int{e^x cos xdx}\)

    \(2\int {{e^x}\cos xdx} \) = \(e^x cosx e^x sin x + C \)

    \(\int {{e^x}\cos xdx} \) = \(\frac{e^x cosx e^x sin x + C}{2} \)