$\begingroup$$$\int\biggl( \sqrt[3]{\frac{\sin x+1}{\cos x}}+\sqrt[3]\frac{\sin x-1}{\cos x}\biggr)\frac{1}{\cos^2x}\,dx ,\;x\in\Bigl(-\frac{\pi}{2},\frac{\pi}{2}\Bigr)$$ 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.
$\endgroup$33 Answers
$\begingroup$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 $$T(s(\tan(x)))-\tan(x)s(\tan(x))$$
$\endgroup$3$\begingroup$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}
$\endgroup$$\begingroup$Hint: Substitute $$\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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