$$\int\biggl( \sqrt[3]{\frac{\sin x+1}{\cos x}}+\sqrt[3]\frac{\sin x-1}{\cos x}\biggr)\frac{1}{\cos^2x}\,dx ,\;x\in\Bigl(-\frac{\pi}{2},\frac{\pi}{2}\Bigr)$$ 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.
Trigonometric integral with tangent
calculusintegration
Related Solutions
The substitution $u=1/x$ yields $dx=-\frac{1}{u^2}\,du$, so the integral becomes $$ \int\frac{-\sin u}{\sqrt{4+3\sin2u}}\,du= \int\frac{-\sin u}{\sqrt{4+3\sin2u}}\,du $$ This can be improved by setting $u=\pi/4-v$, so we get $$ \frac{1}{\sqrt{2}}\int\frac{\cos v-\sin v}{\sqrt{4+3\cos2v}}\,dv= \frac{1}{\sqrt{2}}\biggl( \int\frac{\cos v}{\sqrt{7-6\sin^2v}}\,dv -\int\frac{\sin v}{\sqrt{6\cos^2v+1}}\,dv \biggr) $$ that you should be able to manage.
First of all, recall that $\sqrt{y^2} = \lvert y\rvert,$ not $y$, because $\sqrt{y^2}$ is a non-negative square root even when $y$ is negative. Also note that $0 \leq \sec^{-1}\left(\dfrac x4\right) < \dfrac\pi2$ if and only if $x \geq 4,$ but $\dfrac\pi2 < \sec^{-1}\left(\dfrac x4\right) \leq \pi$ if and only if $x \leq -4.$ Then
\begin{align} \int \frac{\sqrt{16\sec^2\theta - 16}}{4\sec\theta}\, &\cdot 4\sec\theta\tan\theta \,\mathrm d\theta \\ &= \int \sqrt{16\tan^2\theta}\,\tan\theta \,\mathrm d\theta \\ &= 4 \int \lvert\tan\theta\rvert \tan\theta \,\mathrm d\theta \\ &= \begin{cases} \phantom{-}4 \displaystyle\int\tan^2\theta \,\mathrm d\theta & 0 \leq \theta < \dfrac\pi2,\\ -4 \displaystyle\int\tan^2\theta \,\mathrm d\theta &\dfrac\pi2 < \theta \leq \pi \\ \end{cases} \\ &= \begin{cases} 4 \left( \tan\theta - \theta\right) + C_1 & 0 \leq \theta < \dfrac\pi2,\\ 4 \left(\theta - \tan\theta\right) + C_2 & \dfrac\pi2 < \theta \leq \pi \\ \end{cases} \\ &= \begin{cases} 4 \left(\tan \left(\sec^{-1}\left(\dfrac x4\right)\right) - \sec^{-1}\left(\dfrac x4\right)\right) + C_1 & x \geq 4,\\ 4 \left(\sec^{-1}\left(\dfrac x4\right) - \tan \left(\sec^{-1}\left(\dfrac x4\right)\right)\right) + C_2 & x \leq -4. \\ \end{cases} \end{align}
Now when $x \geq 4,$ then $\tan \left(\sec^{-1}\left(\dfrac x4\right)\right) = \dfrac14 \sqrt{x^2 - 16},$ sure enough, but when $x \leq -4$ then $\dfrac\pi2 < \sec^{-1}\left(\dfrac x4\right) \leq \pi$ (second quadrant), hence $\tan \left(\sec^{-1}\left(\dfrac x4\right)\right)$ is negative (or zero), hence $\tan \left(\sec^{-1}\left(\dfrac x4\right)\right) = -\dfrac14 \sqrt{x^2 - 16}.$ So
$$ \int \frac{\sqrt{x^2 - 16}}{x} \, dx = \begin{cases} \sqrt{x^2 - 16} - 4\sec^{-1}\left(\dfrac x4\right) + C_1 & x \geq 4,\\ \sqrt{x^2 - 16} + 4 \sec^{-1}\left(\dfrac x4\right) + C_2 & x \leq -4. \\ \end{cases} $$
Further notice that when $x \geq 4,$ then $\sec^{-1}\left(\dfrac x4\right) = \tan^{-1}\left(\dfrac{\sqrt{x^2-16}}4\right),$ but when $x \leq -4,$ then $\sec^{-1}\left(\dfrac x4\right) = \pi - \tan^{-1}\left(\dfrac{\sqrt{x^2-16}}4\right),$ so the integral can also be written
\begin{align} \int \frac{\sqrt{x^2 - 16}}{x} \, dx &= \begin{cases} \sqrt{x^2 - 16} - 4 \tan^{-1}\left(\dfrac{\sqrt{x^2-16}}4\right) + C_1 & x \geq 4,\\ \sqrt{x^2 - 16} + 4\left(\pi - \tan^{-1}\left(\dfrac{\sqrt{x^2-16}}4\right)\right) + C_2 & x \leq -4 \\ \end{cases} \\ &= \begin{cases} \sqrt{x^2 - 16} - 4 \tan^{-1}\left(\dfrac{\sqrt{x^2-16}}4\right) + C_1 & x \geq 4,\\ \sqrt{x^2 - 16} - 4 \tan^{-1}\left(\dfrac{\sqrt{x^2-16}}4\right) + C_3 & x \leq -4. \\ \end{cases} \\ \end{align}
If we set $C_1 = C_3 = C$ then we have the solution you might get by other means, $$ \int \frac{\sqrt{x^2 - 16}}{x} \, dx = \sqrt{x^2 - 16} - 4 \tan^{-1}\left(\dfrac{\sqrt{x^2-16}}4\right) + C, $$ but in the context of real analysis it is more accurate to say that we can use different constants in the integral for each connected part of the integral's domain.
Some parts of the calculations that were postponed for the sake of the flow of the answer:
Why is $\sec^{-1}\left(\dfrac x4\right) = \tan^{-1}\left(\dfrac{\sqrt{x^2-16}}4\right)$ when $x \geq 4$?
Let $x \geq 4$ and $\sec^{-1}\left(\dfrac x4\right) = \theta.$ Then $\sec\theta = \dfrac x4$ and $\theta$ is in the first quadrant (since $\dfrac x4 > 0$), so
$$ \tan^2\theta = \sec^2\theta - 1 = \dfrac{x^2}{16} - 1 = \dfrac{x^2-16}{16}, $$
so $\tan\theta = \pm \sqrt{\dfrac{x^2-16}{16}}.$ But since $\theta$ is in the first quadrant, $\tan\theta$ must be positive, so $\tan\theta = \sqrt{\dfrac{x^2-16}{16}}$ and (again, since $\theta$ is in the first quadrant) $$\sec^{-1}\left(\dfrac x4\right) = \theta = \tan^{-1}\left(\dfrac{\sqrt{x^2-16}}4\right).$$
Why is $\sec^{-1}\left(\dfrac x4\right) = \pi - \tan^{-1}\left(\dfrac{\sqrt{x^2-16}}4\right)$ when $x \leq -4$?
Note that when $y \leq -1,$ then $\sec^{-1} y = \pi - \sec^{-1}(-y).$ This comes from the way we define the inverse secant function; like the inverse cosine function, we use the second quadrant for negative values. And if $y \leq -1$ then $-y \geq 1,$ so $\sec^{-1}(-y)$ is a first-quadrant angle, so $\pi - \sec^{-1}(-y)$ is a second-quadrant angle and
$$ \sec\left(\pi - \sec^{-1}(-y)\right) = -\sec\left(\sec^{-1}(-y)\right) = -(-y) = y, $$
as required. Now let $x \leq -4$; then $-x \geq 4$ and (substituting $-x$ for $x$ in the what we worked out above for $x \geq 4$) $$ \sec^{-1}\left(\dfrac {(-x)}4\right) = \tan^{-1}\left(\dfrac{\sqrt{(-x)^2-16}}4\right) = \tan^{-1}\left(\dfrac{\sqrt{x^2-16}}4\right), $$ and $$ \sec^{-1}\left(\dfrac x4\right) = \pi - \sec^{-1}\left(\dfrac {(-x)}4\right) = \pi - \tan^{-1}\left(\dfrac{\sqrt{x^2-16}}4\right). $$
Best Answer
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 $$T(s(\tan(x)))-\tan(x)s(\tan(x))$$