[tex]\displaystyle\int\frac{x^3}{\sqrt{x^2+49}}\,\mathrm dx[/tex]
Taking [tex]x=7\tan\theta[/tex] gives [tex]\mathrm dx=7\sec^2\theta\,\mathrm d\theta[/tex], so that the integral becomes
[tex]\displaystyle\int\frac{(7\tan\theta)^3}{\sqrt{(7\tan\theta)^2+49}}(7\sec^2\theta)\,\mathrm d\theta[/tex]
[tex]=\displaystyle7^4\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{49\tan^2\theta+49}}\,\mathrm d\theta[/tex]
[tex]=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\tan^2\theta+1}}\,\mathrm d\theta[/tex]
[tex]=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\sec^2\theta}}\,\mathrm d\theta[/tex]
[tex]=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{|\sec\theta|}\,\mathrm d\theta[/tex]
When [tex]\sec\theta>0[/tex], we have
[tex]=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sec\theta}\,\mathrm d\theta[/tex]
[tex]=\displaystyle7^3\int\tan^3\theta\sec^2\theta\,\mathrm d\theta[/tex]
and from here we can substitute [tex]u=\tan\theta[/tex] to proceed from here.
Quick note: When we set [tex]x=7\tan\theta[/tex], we are implicitly enforcing [tex]-\dfrac\pi2<\theta<\dfrac\pi2[/tex] just so that the substitution can be undone later via [tex]\theta=\tan^{-1}\dfrac x7[/tex]. But note that over this domain, we automatically guarantee that [tex]\sec\theta>0[/tex], so the absolute value bars can be dropped immediately.
