I need to sub $x=\cot^3(\theta)$ in order to integrate: $$\int \sqrt{1+x^{-2/3}}\ dx$$ I know if this were a definite integral, it would change the bounds, but otherwise I'm lost on where to get started. Obviously the first step is $$\int \sqrt{1+\left (\cot^3(\theta)\right )^{-2/3}}\ d\theta$$ but I'm lost as far as a next step.
Substitute x for $\cot^3 \theta$ to evaluate the antiderivative of $\cos^n(x)$
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calculus
integration
trigonometric-integrals
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3If $x=\cot^3(\theta)$, $dx\color{red}{\neq}d\theta$. – 2017-01-27
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0Another way would be to write the integrand as $$\frac{1}{x^{1/3}}\sqrt{1+x^{2/3}}=\frac{3}{2}D(x^{2/3})\cdot\sqrt{1+x^{2/3}}.$$ I think it is now clear what the primitive is. – 2017-01-27
1 Answers
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Making the substitution $x=\cot^3\theta\to\text{d}x=-3\cot^2\theta\csc^2\theta\text{d}\theta$ so the integral is now $-3\int\sec\theta\cot^2\theta\csc^2\theta\text{d}\theta=-3\int\frac{\cos\theta}{\sin^4\theta}\text{d}\theta$ $\overbrace{=}^{u=\sin\theta}-3\int\frac{1}{u^4}\text{d}u=\frac{1}{u^3}+C$ undoing our substitutions we get $$\boxed{\int\sqrt{1+x^{-2/3}}\text{d}x=\left (1+x^{2/3}\right )^{3/2}+C}$$
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0This feels extremely cluttered... I recommend breaking up into multiple lines or something. good nonetheless! – 2017-01-27
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0@BrevanEllefsen I'll fix it for clarity – 2017-01-27