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$$\int\frac{\tan\sqrt{x}}{\sqrt{x}}\ dx$$

I originally thought to let $u=\sqrt{x}$, but then this would mean I have $\displaystyle2\cdot \int\frac{\tan u}{u}\ du$, which is just the same type of problem.

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Actually it would reduce to $$ 2\cdot\int \tan u \,{\rm d}u $$ since ${\rm d}x=2u{\rm d}u$... – AndreasT Feb 8 '13 at 14:56
Be it as it may, multiplying and dividing by 2, without substitution you would have $$ 2\int f\big(g(x)\big)g'(x)\,{\rm d}x = 2F\big(g(x)\big) $$ where $f(x)=\tan x$, $g(x)=\sqrt x$ and $F'=f$. – AndreasT Feb 8 '13 at 14:59
See AndreasT's comment and note $\tan u={\sin u\over \cos u}$. Time for one more substitution... (Note that an appropriate substitution at the start is all that's needed.) – David Mitra Feb 8 '13 at 15:02
@agent54 : You can't just change $dx$ to $du$. That's the most important thing you can learn about how these substitutions work. – Michael Hardy Feb 8 '13 at 15:09
up vote 5 down vote accepted

When you make the substitution you will get $du=\dfrac{1}{2\sqrt{x}}\,dx$ in other words $dx=2u\,du$ Thus the integral becomes$$\int\frac{\tan(u)}{u} 2u\,du=2\int \tan(u)\,du.$$ I think you can continue from here.

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$$u=\sqrt{x}$$ $$du = \frac{1}{2\sqrt{x}}\,dx$$ $$2\,du=\frac{1}{\sqrt{x}}\,dx$$ $$\int\frac{\tan\sqrt{x}}{\sqrt{x}}\,dx = \int(\tan u)\big(2\,du\big)=2\int\tan u\,du$$ $$=2\int\frac{\sin u}{\cos u}\,du = 2\int \frac{1}{\cos u}\Big(-\sin u\,du\Big) = 2\int\frac1w\left(-dw\right)$$

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