Question

integrate tanx/ sqrt(sin^2x-16)

Answer 1

One thing you could do is to multiply the integrand by \(\dfrac{\cos^2x}{\cos^2x}\), so that
\[\begin{align*}\int\frac{\tan x}{\sqrt{\sin^2x-16}}\,\mathrm{d}x&=\int\frac{\sin x\cos x}{\cos^2x\sqrt{\sin^2x-16}}\,\mathrm{d}x\\[1ex]&=\int\frac{\sin x}{\cos^2x\sqrt{\sin^2x-16}}\,\mathrm{d}(\sin x)\end{align*}\]This suggests a substitution of \(u=\sin x\) could work. Now, if \(\sin x=u\), then it follows that \(\cos x=\sqrt{1-u^2}\), so the integral becomes
\[\int\frac{u}{\left(1-u^2\right)\sqrt{u^2-16}}\,\mathrm{d}u\]Another substitution along the lines of either \(t=1-u^2\) or \(t=u^2-16\) should work nicely.

Answer 2

I think!! if you let \(u=\sqrt{sin^2 x-16}\) , then
\(du=\dfrac{sinxcosx}{\sqrt{sin^2x-16}}dx=\dfrac{sinx. cos^2x}{cosx\sqrt{sin^2x-16}}dx\)

Hence \(\dfrac{du}{cos^2x}=\) your integrand.

Now, manipulate \(cos^2x\)

\(u=\sqrt{sin^2 x-16}\\u^2=1-cos^2x-16=-15-cos^2x\\cos^2x = -15-u^2\)

At the end, the whole thing is
\[\int (-15-u^2)du\]