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4.9 Calculating Integral

  1. \(\int x^3 \sqrt{1-x^2} dx\)

    Use a substitution, let \(u = 1 - x^2 \implies du = -2x dx\) and \(x^2 = 1-u \implies -\frac{du}{2} = xdx\)

    Then \(\int x^2\sqrt{1-x^2}x dx = \int (1-u)\sqrt{u}(-\frac{1}{2})du\) \(= -\frac{1}{2}\int (\sqrt{u}-u^{\frac{3}{2}})du\)

    \(= -\frac{1}{2}(\frac{2}{3}u^{\frac{3}{2}} - \frac{2}{5}u^{\frac{5}{2}}) + C\) \(= -\frac{1}{2}(\frac{2}{3}(1-x^2)^{\frac{3}{2}} - \frac{2}{5}(1-x^2)^{\frac{5}{2}}) + C\)

  2. \(\int \sin(x) \cos(x) dx\)

    Let \(u = \sin(x) \implies du = \cos(x) dx\), then \(\int u du = \frac{u^2}{2} + C = \frac{\sin^2(x)}{2} + C\)

    In general \(\int \sin^n(x) \cos(x) dx\), then \(u = \sin(x) \implies du = \cos(x) dx\)

    Thus \(\int u^n du = \frac{u^{n+1}}{n+1} + C = \frac{\sin^{n+1}(x)}{n+1} + C\)

    Similarly, \(\int \cos^{n}(x) \sin(x) dx\)

    Let \(u = \cos(x) \implies du = -\sin(x)dx\)

    Then \(\int\cos^{n}(x)\sin(x)dx=-\frac{\cos^{n+1}(x)}{n+1}+C\)

  3. \(\int \sin^3(x) dx = \int \sin^2(x) \sin(x) dx\)

    \(= \int (1-\cos^2(x))\sin(x) dx\)

    \(= \int \sin(x)dx - \int \cos^2(x) \sin(x) dx\)

    \(= -\cos(x) + \frac{\cos^3(x)}{3} + C\)

    Similarly, \(\int \cos^3(x) dx = \sin(x) - \frac{\sin^3(x)}{3} + C\)

  4. \(\int \sin^2(x) dx = \int (\frac{1}{2} - \frac{\cos(2x)}{2}) dx\)

    \(= \frac{1}{2}\int dx - \frac{1}{2}\int \cos(2x) dx\)

    Let \(u = 2x, du = 2 dx\)

    \(= \frac{1}{2}x - \frac{1}{2} \int \cos(u) \frac{du}{2}\) \(= \frac{x}{2} - \frac{1}{4} \int \cos(u) du = \frac{x}{2} - \frac{1}{4} \sin(u) + C\)

    \(= \frac{x}{2} - \frac{1}{4} \sin(2x) + C\)

  5. \(\int \sec(x) dx\)

    \(= \int \sec(x) \frac{\sec(x) + \tan(x)}{\sec(x) + \tan(x)} dx\) \(= \int \frac{\sec^2(x) + \sec(x) \tan(x)}{\sec(x) + \tan(x)} dx\)

    Substitution: \(u = \sec(x) + \tan(x)\), then \(du = (\tan(x)\sec(x) + \sec^2(x)) dx\)

    \(= \int \frac{1}{u} du\) \(= \ln(|u|) + C = \ln(|\sec(x) + \tan(x)|) + C\)

    Because \(\tan(x) = \frac{\sin(x)}{\cos(x)}\) and \(\sec(x) = \frac{1}{\cos(x)}\)

    Then \(\frac{d}{dx} \tan(x) = \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)}\) \(= \sec^2(x)\) and \(\frac{d}{dx} \sec(x) = \frac{d}{dx} (\cos^{-1}(x)) = \frac{-\sin(x)}{-\cos^2(x)}\) \(= \tan(x) \cdot \sec(x)\)

  6. \(\int x e^x dx\)

    We want to use integration by parts. How do we choose \(u\) and \(v'\)?
    We pick \(u\) in the following order:
    (1) inverse of trigonometric functions \(\arcsin, \arccos,...\)​
    (2) logarithmic function \(\ln(x)\)
    (3) algebraic function \(x^n, \sqrt{x}, ...\)
    (4) trigonometric function \(\sin(x), \cos(x), ...\)
    (5) exponential functions \(e^x, 2^x, ...\)

    In our case \(u(x)=x \implies du = dx\) and \(v'(x) dx = e^x dx \implies v = \int e^x dx = e^x\)

    By parts \(\int x e^x dx = x e^x - \int e^x dx\) \(= xe^x - e^x + C\)

  7. \(\int \sec^3(x) dx = \int \sec(x) \sec^2(x) dx\)

    Let \(u = \sec(x) \implies du = \tan(x)\sec(x) dx\)
    ​\(v' = \sec^2(x) \implies v = \tan(x)\)

    Now, by parts

    \(\int \sec^3(x) dx = \sec(x)\tan(x) - \int \tan(x) (\tan(x)\sec(x)) dx\)
    ​\(= \sec(x)\tan(x) - \int \tan^2(x) \sec(x) dx\)
    ​\(= \sec(x)\tan(x) - \int (\sec^3(x) - \sec(x)) dx\)
    ​\(= \sec(x)\tan(x) - \int \sec^3(x) dx + \int \sec(x) dx\)

    So \(\int \sec^{3}(x) dx = \frac{1}{2}(\sec(x)\tan(x) + \ln|\sec(x) + \tan(x)|) + C\)

    Because \(\tan^2(x)\sec(x) = \frac{\sin^2(x)}{\cos^2(x)} \sec(x) = \frac{(1-\cos^2(x))}{\cos^2(x)} \sec(x) = (\sec^2(x) - 1) \sec(x)\)

  8. \(\int e^x \sin(x) dx\). We want to integrate by parts.

    Let's \(u = \sin(x) \implies du = \cos(x) dx\) and \(v' = e^x \implies v = e^x\)

    Now, by parts \(\int e^{x} \sin(x) dx = \sin(x)e^{x} - \int e^{x} \cos(x) dx\)

    Solve \(\int e^x \cos(x) dx\), we use parts again
    Let's \(u = \cos(x) \implies du = -\sin(x) dx\) and \(v' = e^x \implies v = e^x\)

    Thus \(\int e^x \cos(x) dx = \cos(x)e^x - \int e^x (-\sin(x) dx)\) \(= \cos(x) e^x + \int e^x \sin(x) dx\)

    Then \(\int e^{x}\sin(x)dx=\sin(x)e^{x}-(\cos(x)e^{x}+\int e^{x}\sin(x)dx)\)

    Then \(\int e^x \sin(x) dx = \frac{1}{2} (\sin(x) - \cos(x))e^x + C\)

  9. \(\int \frac{4x-11}{x^3 - 9x^2} dx = \int \frac{4x-11}{x^2(x-9)} dx\)
    We use partial fractions.

    \(\frac{4x-11}{x^2(x-9)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-9} = \frac{Ax(x-9)+B(x-9)+Cx^2}{x^2(x-9)}\)

    Then \(4x-11=Ax(x-9)+B(x-9)+Cx^{2}\)

    Give some value to x
    ​\(x = 0 \implies -11 = -9B \implies B = \frac{11}{9}\)
    ​\(x = 9 \implies 36 - 11 = C \cdot 81 \implies 25 = 81C \implies C = \frac{25}{81}\)

  10. \(\int \frac{x^2 + 2x + 3}{(x-6)(x^2+4)} dx\)

    \(\frac{x^2 + 2x + 3}{(x-6)(x^2+4)} = \frac{A}{x-6} + \frac{Bx + C}{x^2+4} = \frac{A(x^2+4) + (Bx+C)(x-6)}{(x-6)(x^2+4)} = \frac{Ax^2 + 4A + Bx^2 - 6Bx + Cx - 6C}{(x-6)(x^2+4)}\)

    Then \(x^2 + 2x + 3 = (A+B)x^2 + (-6B+C)x + (4A - 6C)\)image

  11. image

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\(\int \tan x \, dx = -\ln|\cos x| + C\)​

\(\int \cot x \, dx = \ln|\sin x| + C\)

\(\int \sec x \, dx = \ln|\sec x + \tan x| + C\)

\(\int \csc x \, dx = \ln|\csc x - \cot x| + C\)​

\(\int \sec^2 x \, dx = \tan x + C\)

\(\int \csc^2 x \, dx = -\cot x + C\)

\(\int \frac{1}{x} \, dx = \ln|x| + C\)

\(\int \frac{1}{1+x^2} \, dx = \arctan x + C\)​

\(\int \frac{1}{\sqrt{1-x^2}} \, dx = \arcsin x + C\)​