Indefinite integration (9.3-9.6)

Use integration by substitution to find indefinite integrals

  • Theory
  • Examples
    Find \(\int {\,\csc 3x{{\cot }^2}3x\,dx} \).
  • Solutions
    \(\begin{array}{1}\int {\,\csc 3x{{\cot }^2}3x\,dx} = - {\rm{ }}\int {\,\cot 3x \cdot ( - \cot 3x\csc 3x)\,dx} \\ = - {\rm{ }}\frac{1}{3}\int {\,\cot 3x\,d(\csc 3x)} \\ = - {\rm{ }}\frac{1}{3}\cot 3x\csc 3x + \frac{1}{3}\int {\,\csc 3x\,d(\cot 3x)} \\ = - {\rm{ }}\frac{1}{3}\cot 3x\csc 3x - \int {\,{{\csc }^3}3x\,dx} \\ = - {\rm{ }}\frac{1}{3}\cot 3x\csc 3x - \int {\,({{\cot }^2}3x + 1)\csc 3x\,dx} \\ = - {\rm{ }}\frac{1}{3}\cot 3x\csc 3x - \int {\,{{\cot }^2}3x\csc 3x\,dx} - \int {\,\csc 3x\,dx} \end{array}\)

    \(\begin{array}{1}2\int {\,\csc 3x{{\cot }^2}3x\,dx} = - {\rm{ }}\frac{1}{3}\cot 3x\csc 3x - \int {\,\csc 3x\,dx} \\
    \int {\,\csc 3x{{\cot }^2}3x\,dx} = - {\rm{ }}\frac{1}{6}\cot 3x\csc 3x - \frac{1}{2}\int {\,\csc 3x\,dx} \\ = \underline{\underline { - {\rm{ }}\frac{1}{6}\cot 3x\csc 3x - \frac{1}{6}\ln \left| {{\kern 1pt} \csc 3x - \cot 3x{\kern 1pt} } \right| + C}} \end{array}\)

Use trigonometric substitutions to find the indefinite integrals involving [FormalMissing]

  • Theory
  • Examples
    Find \(\int {\,\frac{{{x^3}}}{{\sqrt {4{x^2} - 25} }}dx} \).
  • Solutions
    Let \(x = \frac{5}{2}\sec \theta \) where (\(0 < \theta < \frac{\pi }{2}\) or\(\pi < \theta < \frac{{3\pi }}{2}\)),
    then\(dx = \frac{5}{2}\tan \theta \sec \theta \,d\theta \) and\(\sqrt {4{x^2} - 25} = 5\tan \theta \).
    [GraphMissing M2E09 Q198]
    \(\begin{array}{1}\int {\,\frac{{{x^3}}}{{\sqrt {4{x^2} - 25} }}dx} = \int {\,\frac{{{\textstyle{{125} \over 8}}{{\sec }^3}\theta }}{{5\tan \theta }} \cdot \frac{5}{2}\tan \theta \sec \theta \,d\theta } \\ = \frac{{125}}{{16}}\int {\,{{\sec }^4}\theta \,d\theta } \\ = \frac{{125}}{{16}}\int {\,({{\tan }^2}\theta + 1) \cdot {{\sec }^2}\theta \,d\theta } \\ = \frac{{125}}{{16}}\int {\,({{\tan }^2}\theta + 1)} \,d(\tan \theta )\\ = \frac{{125}}{{16}}{\rm{(}}\frac{{{{\tan }^3}\theta }}{3} + \tan \theta ) + C\\ = \frac{{125}}{{48}} \cdot \frac{{{{(4{x^2} - 25)}^{\frac{3}{2}}}}}{{125}} + \frac{{125}}{{16}} \cdot \frac{{{{(4{x^2} - 25)}^{\frac{1}{2}}}}}{5} + C\\ = \underline{\underline {\frac{1}{{48}}{{(4{x^2} - 25)}^{\frac{3}{2}}} + \frac{{25}}{{16}}{{(4{x^2} - 25)}^{\frac{1}{2}}} + C}} \end{array}\)

Use integration by parts to find indefinite integrals

  • Theory
  • Examples
    (a) Find \(\int {\,{e^{ - x}}\sin 2x\,dx} \).
    (b) Hence find \(\int {\,{e^{ - x}}{{(\sin x + \cos x)}^2}dx} \).
  • Solutions
    (a)
    \(\begin{array}{1}\int {\,{e^{ - x}}\sin 2x\,dx} = - {\rm{ }}\int {\,\sin 2x\,d({e^{ - x}})} \\
    = - {\rm{ }}{e^{ - x}}\sin 2x + \int {\,{e^{ - x}}d(\sin 2x)} \\
    = - {\rm{ }}{e^{ - x}}\sin 2x + 2\int {\,{e^{ - x}}\cos 2x\,dx} \\ \int {\,{e^{ - x}}\cos 2x\,dx} = - {\rm{ }}\int {\,\cos 2x\,d({e^{ - x}})} \\
    = - {\rm{ }}{e^{ - x}}\cos 2x + \int {\,{e^{ - x}}d(\cos 2x)} \\
    = - {\rm{ }}{e^{ - x}}\cos 2x - 2\int {\,{e^{ - x}}\sin 2x\,dx} \end{array}\)

    \(\begin{array}{1}\int {\,{e^{ - x}}\sin 2x\,dx} = - {\rm{ }}{e^{ - x}}\sin 2x + 2( - {e^{ - x}}\cos 2x - 2\int {\,{e^{ - x}}\sin 2x\,dx} )\\
    5\int {\,{e^{ - x}}\sin 2x\,dx} = - {\rm{ }}{e^{ - x}}\sin 2x - 2{e^{ - x}}\cos 2x + {C_1}^{^{}}\\
    \int {{e^{ - x}}\sin 2x\,dx} = \underline{\underline { - {\rm{ }}\frac{{{e^{ - x}}}}{5}(\sin 2x + 2\cos 2x) + C}} \end{array}\)

    (b)
    \(\begin{array}{1}\int {\,{e^{ - x}}{{(\sin x + \cos x)}^2}dx} = \int {\,{e^{ - x}}({{\sin }^2}x + 2\sin x\cos x + {{\cos }^2}x)\,dx} \\
    = \int {\,{e^{ - x}}(1 + \sin 2x)\,dx} \\
    = \int {\,{e^{ - x}}dx} + \int {\,{e^{ - x}}\sin 2x\,dx} \\
    = \underline{\underline { - {\rm{ }}{e^{ - x}} - \frac{{{e^{ - x}}}}{5}(\sin 2x + 2\cos 2x) + C}} \end{array}\)