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$$\int\biggl( \sqrt[3]{\frac{\sin x+1}{\cos x}}+\sqrt[3]\frac{\sin x-1}{\cos x}\biggr)\frac{1}{\cos^2x}\,dx ,\;x\in\Bigl(-\frac{\pi}{2},\frac{\pi}{2}\Bigr)$$ Can somebody give some tips about how can I evaluate this integral,please? I observed that $\frac{1}{\cos^2x}$ is the derivative of tangent, but I don't know how to do it using the substitution $t=\tan x$ because there will be $\frac{1}{\cos x}$ inside and I think there is another method.

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  • $\begingroup$ Are you trying to evaluate the integral with the limits $\int_{-\pi/2}^{\pi/2}$, or the indefinite integral? $\endgroup$ – John Doe Jan 12 at 20:09
  • $\begingroup$ Maybe it should be $\frac{1+\sin{x}}{\cos^2x}$ instead $\frac{1}{\cos^2x}$? $\endgroup$ – Michael Rozenberg Jan 12 at 20:11
  • $\begingroup$ Notice, that, when you substitute $x + \pi$ for $x$ in the integrand, you get the same function, so according to wikipedia, a sensible change of variable will be $\tan$. $\endgroup$ – Viktor Glombik Jan 12 at 20:17
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I will use integration by parts twice. First we write \begin{align} I &= \int \left (\sqrt[3]{\frac{\sin x + 1}{\cos x}} + \sqrt[3]{\frac{\sin x - 1}{\cos x}} \right ) \sec^2 x \, dx\\ &= \int \big{(}\sqrt[3]{\tan x + \sec x} + \sqrt[3]{\tan x - \sec x} \big{)} \sec^2 x \, dx. \end{align}

If we let $$f(x) = \sqrt[3]{\tan x + \sec x} + \sqrt[3]{\tan x - \sec x}$$ and $$g(x) = \sqrt[3]{\tan x + \sec x} - \sqrt[3]{\tan x - \sec x},$$ then obsereve that $$f'(x) = \frac{1}{3} \sec x \cdot g(x) \quad \text{and} \quad g'(x) = \frac{1}{3} \sec x \cdot f(x).$$ So for the integral we have \begin{align} I &= \int \sec^2 x \cdot f(x) \, dx\\ &= f(x) \cdot \tan x - \int \tan x \cdot f'(x) \, dx \qquad \text{(by parts)}\\ &= f(x) \cdot \tan x - \frac{1}{3} \int \sec x \tan x \cdot g(x) \, dx \\ &= f(x) \cdot \tan x - \frac{1}{3} \sec x \cdot g(x) + \frac{1}{3} \int \sec x \cdot g'(x) \, dx \quad \text{(by parts)}\\ &= f(x) \cdot \tan x - \frac{1}{3} \sec x \cdot g(x) + \frac{1}{9} \int \sec^2 x \cdot f(x) \, dx\\ &= f(x) \cdot \tan x - \frac{1}{3} \sec x \cdot g(x) + \frac{1}{9} I, \end{align} giving $$I = \frac{9}{8} \tan x \cdot f(x) - \frac{3}{8} \sec x \cdot g(x) + C$$ or \begin{align} I &= \frac{9}{8} \tan x \big{(} \sqrt[3]{\tan x + \sec x} + \sqrt[3]{\tan x - \sec x} \big{)}\\ & \qquad - \frac{3}{8} \sec x \big{(} \sqrt[3]{\tan x + \sec x} - \sqrt[3]{\tan x - \sec x} \big{)} + C. \end{align}

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First note that if $f$ is a function with antiderivative $F$, then the antiderivative of its inverse function $f^{-1}(x)$ is $F( f^{-1}(x))-xf^{-1}(x)$, which can be easily shown by differentiating.

Now let's consider your integral.

Making a trigonometric substitution allows us to consider instead the following integrand: $$s(x)=\sqrt[3]{x+\sqrt{x^2+1}}+\sqrt[3]{x-\sqrt{x^2+1}}$$ Notice that $s$ satisfies the property $$s=\sqrt[3]{2x-3s}$$ or $$s^3+3s=2x$$ If $t$ is the inverse function of $s$ so that $s(t(x))=x$ and $t(s(x))=x$, we have from the above equation that $$2t=x^3+3x$$ Thus, we may use the formula mentioned at the beginning of the post to solve your integral. The antiderivative of $t$ with respect to $x$ is $$T(x)=\frac{x^4+6x^2}{8}$$ and so the antiderivative of $s$ is given by $$S(x)=T(s(x))-xs(x)$$ where $T$ and $s$ are defined above. This yields the desired result, and now all you have to do is undo the trig substitution at the beginning, and the value of your integral is $$T(s(\tan(x)))-\tan(x)s(\tan(x))$$

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  • $\begingroup$ Is the approach taken here a known method to approach a certain family(ies) of integrals? (similar to say the Half Tangent substitution method for rational trigonometric functions) $\endgroup$ – user150203 Jan 13 at 0:00
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    $\begingroup$ @DavidG Not that I know of. I just worked it out from scratch. $\endgroup$ – Franklin Pezzuti Dyer Jan 13 at 0:36
  • $\begingroup$ How did you get that expresion of s(x) ? $\endgroup$ – Gaboru Jan 13 at 1:36
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Hint: Substitute $$\sin(x)=\frac{2t}{1+t^2}$$ $$\cos(x)=\frac{1-t^2}{1+t^2}$$ $$dx=\frac{2dt}{1+t^2}$$

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    $\begingroup$ NONONONONO it makes things wayyyyy worse $\endgroup$ – clathratus Jan 12 at 20:49
  • $\begingroup$ This gives $$\int\left(\sqrt[3]\frac{1+t}{1-t}+\sqrt[3]\frac{t-1}{t+1}\right)\frac{2(1+t^2)}{(1-t^2)^2}\,dt$$...what then? $\endgroup$ – John Doe Jan 12 at 21:27
  • $\begingroup$ Now i would substitute the roots. $\endgroup$ – Dr. Sonnhard Graubner Jan 12 at 21:35

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