Some Standard Integrals - SS3 Mathematics Lesson Note
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\(\frac{d}{dx}\left( ax^{n} \right) = anx^{n - 1}\ \ \ \ \ \ \ \therefore\ \int_{}^{}{ax^{n}}\ dx = \ \frac{ax^{n + 1}}{n + 1} + C\ (provided\ n \neq - 1)\)
\(\frac{d}{dx}\left( \sin x \right) = \cos{x\ \ \ \ \ \ \ \ \ \ \therefore}\ \int_{}^{}{\cos x}\ dx = \sin x + C\)
\(\frac{d}{dx}\left( \sin{ax} \right) = \ \ a\cos{ax\ \therefore\ \int_{}^{}{\cos{ax}}}\ dx\ = \frac{\sin{ax}}{a} + C\)
\(\frac{d}{dx}\left( \cos x \right) = \ - \sin{x\ \ \ \ \ \ \ \ \therefore\int_{}^{}{\sin x}}\ dx = \ - \cos x + C\)
\(\frac{d}{dx}\left( \cos{ax} \right) = - a\sin{ax}\ \ \ \therefore\ \int_{}^{}{\sin{ax}}\ dx = \ - \frac{\cos{ax}}{a} + C\)
\(\frac{d}{dx}\left( e^{x} \right) = e^{x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore\ \int_{}^{}e^{x}\ dx = e^{x} + C\)
\(\frac{d}{dx}\left( e^{ax} \right) = ae^{ax}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore\ \int_{}^{}e^{ax}\ dx = \frac{e^{ax}}{a} + C\)
\(\frac{d}{dx}\left( a^{x} \right) = a^{x}\ln{a\ \ \ \ \ \ \ \ \ \ \ \ \ \therefore}\ \int_{}^{}a^{x}\ dx = \frac{a^{x}}{\ln a} + C,\ where\ln{a = \ \log_{e}a}\)
\(\frac{d}{dx}(\ln{x)} = \frac{1}{x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore\ \int_{}^{}{\frac{1}{x}\ }dx = \ln{x + C}\)
\(\frac{d}{dx}(\tan{x)} = \sec^{2}x\ \ \ \ \ \ \ \ \ \ \therefore\ \int_{}^{}{\sec^{2}x}\ dx = \tan{x + C}\)
\(\frac{d}{dx}\left( \cot x \right) = \ - cosec^{2}x\ \ \ \ \therefore\ \int_{}^{}{{cosec}^{2}x}\ dx = \ - \cot x + C\)
Example: Integrate the following with respect to \(x\): (a) \(6x^{4}\) (b) \(\sin{5x}\) (c) \(\frac{5}{x}\)
Solution:
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\(\int_{}^{}{6x^{4}}\ dx = 6\int_{}^{}x^{4}\ dx = 6.\frac{x^{5}}{5} + C = \frac{6x^{5}}{5} + C\)
\(\int_{}^{}{\sin{5x}}\ dx = \ - \frac{\cos{5x}}{5} + C = - \frac{1}{5}\cos{5x} + C\)
\(\int_{}^{}\frac{5}{x}\ dx = 5\int_{}^{}\frac{1}{x}\ dx = 5\ln x + C\)