I have the integral $$\int\frac{\sqrt{16x^2-9}}x\,\mathrm{d}x,$$ and I am having trouble doing the trigonometric substitution. So for integrals in the from of $\sqrt{x^2-a^2}$ where $a$ is a constant is by substituting $x$ for $a\sec\theta$. This is where I am having trouble making this substitution. I will post the steps I have taken and someone please point me in the right direction, thanks a lot for the help in advance. $$\int\frac{\sqrt{16x^2-9}}x\,\mathrm{d}x$$ $$a=3$$ $$x=3\sec\theta$$ $$\mathrm{d}x=3\sec\theta\tan\theta\,\mathrm{d}\theta$$ so $$16x^2-9\rightarrow 16\left(3\sec\theta^2-9\right)$$ $$144\sec^2\theta-9 \rightarrow 144\left(\sec^2\theta-\frac1{16}\right)$$ and that is where I would normally substitute that into $\tan^2\theta$ but I am not getting how to go about doing that. Thanks again for all the help in advance again.
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2 Answers
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we have $$\int {\sqrt{16x^2-9}\over x} dx=\int {x\sqrt{16x^2-9}\over x^2} dx$$ now Put $16x^2-9=u^2$ Therefore, we have $$\int {x\sqrt{16x^2-9}\over x^2} dx=\int\frac{u^2}{u^2+9}du$$ I think from here you can do
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Hint:
$$\int\frac{\sqrt{16x^2-9}}{x}dx=\int\frac{\sqrt{(4x)^2-9}}{x}dx$$
you can make $v=4x,dv=4dx$ then you get
$$\int\frac{\sqrt{v^2-9}}{v}dv$$
now you can solve the integral above with the substituition that you maked and back to the variable $x$ making $v=4x$.
