I am quite stuck on how do I solve this indefinite integral, the integrand is: (tan(x))^2(sec(x))^1/2; I got it as a residue of a differential equation I was solving, I tried a lot of substitutions but they only make it more and more complicated, I tried to convert tangent into secant to obtain a better integral but it only gets more and more complicated.
Apologies for I uploaded the integrand in this manner for I am still a beginner and learning how to use Latex, or MathJax.
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Hint:
Find out that $$\int_0^x\tan^2(t)\sec^{1/2}(t)dt=\int\tan^2(x)\sec^{1/2}(x)dx$$
Then express everything in the integrand in terms of $\sin x$ and $\cos x$ raised to powers, then apply $$\int_0^x\sin^{a}(t)\cos^{b}(t)dt=\frac12B\bigg(\sin^2x;\frac{a+1}{2},\frac{b+1}{2}\bigg)$$ Where $B(x;h,k)$ is the Incomplete Beta function.
EDIT:
Here's how you derive that identity I gave you. $$I(x,a,b)=\int_0^x\sin^{a}(t)\cos^{b}(t)dt$$ The substitution $u=\sin^2(t)$ gives $$I(x,a,b)=\frac12\int_{0}^{\sin^2x}u^{\frac{a-1}2}(1-u)^{\frac{b-1}2}du$$ $$I(x,a,b)=\frac12\int_{0}^{\sin^2x}u^{\frac{a+1}2-1}(1-u)^{\frac{b+1}2-1}du$$ Then one must recall the primary definition for the incomplete Beta function $$B(x;h,k)=\int_0^xt^{h-1}(1-t)^{k-1}dt$$ Of course this gives our integral: $$I(x,a,b)=\frac12B\bigg(\sin^2x;\frac{a+1}2,\frac{b+1}2\bigg)$$
Maple and Wolfram Alpha agree that the antiderivative involves elliptic integrals. Elementary techniques will not work.