# Trigonometry Everywhere - 1

$I=\int x^3 dx$ $let \space x=\red{\cos\theta}\Rightarrow dx=-\sin\theta d\theta$ $\Rightarrow I=-\int\cos^3\theta\sin\theta d\theta$ $=\dfrac{-1}{2}\int (\cos 2\theta\cos\theta+\cos\theta)\sin\theta d\theta$ $=\dfrac{-1}{2}\int\cos 2\theta\sin\theta\cos\theta d\theta - \dfrac{1}{2}\int\cos\theta\sin\theta d\theta$ $=\dfrac{-1}{4}\int\cos 2\theta\sin 2\theta d\theta - \dfrac{1}{4}\int\sin 2\theta d\theta$ $= \dfrac{-1}{8}\int\sin 4\theta d\theta - \dfrac{1}{4}\int\sin 2\theta d\theta$ $=\dfrac{-1}{8}\times\dfrac{-\cos 4\theta}{4} - \dfrac{1}{4}\times\dfrac{-\cos 2\theta}{2}+c$ $=\dfrac{\cos 4\theta}{32}+\dfrac{\cos 2\theta}{8}+c$ $=\dfrac{2\cos^2 2\theta - 1}{32}+\dfrac{2\cos^2\theta-1}{8}+c$ $=\dfrac{2(2\red{\cos^2\theta}-1)^2 - 1}{32}+\dfrac{2\red{\cos^2\theta}-1}{8}+c$ $=\dfrac{2(2\red{x^2}-1)^2 - 1}{32}+\dfrac{2\red{x^2}-1}{8}+c$ $=\dfrac{2(4x^4+1-4x^2) - 1}{32}+\dfrac{2x^2{-1}}{8}\blue{+c}$ $= \dfrac{8x^4\blue{+2}-8x^2 \blue{-1}}{32}+\dfrac{2x^2\blue{-1}}{8}+c$ $= \red{\dfrac{8x^4}{32}}\blue{+\dfrac{1}{32}}\red{-\dfrac{8x^2}{32}+\dfrac{2x^2}{8}}\blue{-\dfrac{1}{8}+c}$ $= \red{\dfrac{x^4}{4}}\cancel{\red{-\dfrac{x^2}{4}}}\cancel{\red{+\dfrac{x^2}{4}}}+\blue{C}$ $= \red{\dfrac{x^4}{4}}+\blue{C}$

$Inspiration$

Note by Zakir Husain
1 month ago

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What about hyperbolic trigonometric substitution: $x = \sinh \theta$?

- 1 month ago

$I=\int x^3dx$ $let\space x=\red{\sinh\theta}\Rightarrow dx=\cosh\theta d\theta$ $\Rightarrow I=\int\sinh^3\theta\cosh\theta d\theta$ $=\dfrac{1}{2}\int\dfrac{\sinh^2\theta\sinh 2\theta}{\cancel{\red{\cosh\theta}}}\cancel{\red{\cosh\theta}} d\theta$ $=\dfrac{1}{2}\int\sinh 2\theta\sinh^2\theta d\theta$ $=\dfrac{1}{4}\int\sinh 2\theta(\cosh 2\theta - 1) d\theta$ $=\dfrac{1}{4}\int\sinh 2\theta\cosh 2\theta d\theta - \dfrac{1}{4}\int\sinh 2\theta d\theta$ $=\dfrac{1}{8}\int\sinh 4\theta d\theta - \dfrac{1}{4}\int\sinh 2\theta d\theta$ $=\dfrac{1}{8}\times\dfrac{\cosh 4\theta}{4} - \dfrac{1}{4}\times\dfrac{\cosh 2\theta}{2}+\blue{c}$ $=\dfrac{2\cosh^2\theta-1}{32} - \dfrac{2{\sinh^2\theta}+1}{8}+\blue{c}$ $=\dfrac{2(2\red{\sinh\theta}^2+1)^2-1}{32} - \dfrac{2\red{\sinh^2\theta}+1}{8}+\blue{c}$ $=\dfrac{2(2\red{x}^2+1)^2\blue{-1}}{32} - \dfrac{2\red{x}^2\blue{+1}}{8}+\blue{c}$ $=\dfrac{8x^4+8x^2\blue{+2-1}}{32} - \dfrac{2\red{x}^2\blue{+1}}{8}+\blue{c}$ $=\red{\dfrac{8x^4}{32}}+\red{\dfrac{8x^2}{32}}+\blue{\dfrac{{1}}{32}} - \red{\dfrac{2{x}^2}{8}}-\blue{\dfrac{{1}}{8}}+\blue{c}$ $=\red{\dfrac{x^4}{4}}+\cancel{\red{\dfrac{x^2}{4}}}- \cancel{\red{\dfrac{{x}^2}{4}}}+\blue{C}$ $=\red{\dfrac{x^4}{4}}+\blue{C}$

- 1 month ago

Beautiful. Now what about integration by parts? $\int x^3 \, dx = \int x \cdot x^2 \, dx$?

;) ;) ;) ;) ;) ;) ;) ;) ;) ;) ;) ;) ;) ;) ;) ;) ;) ;)

- 1 month ago

It's not much interesting :(

- 1 month ago