Trigonometry Everywhere - 1

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

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#Calculus

Note by Zakir Husain
2 months, 3 weeks ago

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Comments

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

Pi Han Goh - 2 months, 3 weeks ago

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I=∫x3dxI=\int x^3dx let x=sinh⁡θ⇒dx=cosh⁡θdθlet\space x=\red{\sinh\theta}\Rightarrow dx=\cosh\theta d\theta ⇒I=∫sinh⁡3θcosh⁡θdθ\Rightarrow I=\int\sinh^3\theta\cosh\theta d\theta =12∫sinh⁡2θsinh⁡2θcosh⁡θcosh⁡θdθ=\dfrac{1}{2}\int\dfrac{\sinh^2\theta\sinh 2\theta}{\cancel{\red{\cosh\theta}}}\cancel{\red{\cosh\theta}} d\theta =12∫sinh⁡2θsinh⁡2θdθ=\dfrac{1}{2}\int\sinh 2\theta\sinh^2\theta d\theta =14∫sinh⁡2θ(cosh⁡2θ−1)dθ=\dfrac{1}{4}\int\sinh 2\theta(\cosh 2\theta - 1) d\theta =14∫sinh⁡2θcosh⁡2θdθ−14∫sinh⁡2θdθ=\dfrac{1}{4}\int\sinh 2\theta\cosh 2\theta d\theta - \dfrac{1}{4}\int\sinh 2\theta d\theta =18∫sinh⁡4θdθ−14∫sinh⁡2θdθ=\dfrac{1}{8}\int\sinh 4\theta d\theta - \dfrac{1}{4}\int\sinh 2\theta d\theta =18×cosh⁡4θ4−14×cosh⁡2θ2+c=\dfrac{1}{8}\times\dfrac{\cosh 4\theta}{4} - \dfrac{1}{4}\times\dfrac{\cosh 2\theta}{2}+\blue{c} =2cosh⁡2θ−132−2sinh⁡2θ+18+c=\dfrac{2\cosh^2\theta-1}{32} - \dfrac{2{\sinh^2\theta}+1}{8}+\blue{c} =2(2sinh⁡θ2+1)2−132−2sinh⁡2θ+18+c=\dfrac{2(2\red{\sinh\theta}^2+1)^2-1}{32} - \dfrac{2\red{\sinh^2\theta}+1}{8}+\blue{c} =2(2x2+1)2−132−2x2+18+c=\dfrac{2(2\red{x}^2+1)^2\blue{-1}}{32} - \dfrac{2\red{x}^2\blue{+1}}{8}+\blue{c} =8x4+8x2+2−132−2x2+18+c=\dfrac{8x^4+8x^2\blue{+2-1}}{32} - \dfrac{2\red{x}^2\blue{+1}}{8}+\blue{c} =8x432+8x232+132−2x28−18+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} =x44+x24−x24+C=\red{\dfrac{x^4}{4}}+\cancel{\red{\dfrac{x^2}{4}}}- \cancel{\red{\dfrac{{x}^2}{4}}}+\blue{C} =x44+C=\red{\dfrac{x^4}{4}}+\blue{C}

Zakir Husain - 2 months, 3 weeks ago

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Beautiful. Now what about integration by parts? ∫x3 dx=∫x⋅x2 dx\int x^3 \, dx = \int x \cdot x^2 \, dx ?

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

Pi Han Goh - 2 months, 3 weeks ago

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@Pi Han Goh – It's not much interesting :(

Zakir Husain - 2 months, 3 weeks ago
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