I need help with this integral: $$\int\frac{\sqrt{\tan x}}{\cos^2x}dx$$ I tried substitution and other methods, but all have lead me to this expression: $$2\int\sqrt{\tan x}(1+\tan^2 x)dx$$ where I can't calculate anything... Any suggestions? Thanks!
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8$\begingroup$ Hint: what's the derivative of $\tan x$? $\endgroup$– Simon SCommented Feb 9, 2015 at 19:55
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$\begingroup$ Hint: rewrite as $$\int \sec^2 x\sqrt{\tan x}dx$$ $\endgroup$– John JoyCommented Feb 10, 2015 at 1:09
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$\begingroup$ This forum would be so much better if people only offered hints like you have done. Much more beneficial than full solutions. $\endgroup$– mmgro27Commented Aug 13, 2015 at 14:07
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4 Answers
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Hint: Let $u = \tan x$ then $du = \sec^2 x\ \ dx = \frac{1}{\cos^2 x} dx$
answered Feb 9, 2015 at 19:59
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$$ \int \dfrac{\sqrt{\tan x}}{\cos^2 x} dx = \int \sqrt{\tan x} \; d(\tan x) = \dfrac{2}{3}\sqrt{(\tan x)^3} +C $$
answered Feb 9, 2015 at 20:05
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Set $t=\tan x$; then $\,\mathrm d\mkern1.5mu t=\dfrac1{\cos^2x}\mathrm d\mkern1.5mu x$ so the integral becomes $$\int\sqrt t\,\mathrm d\mkern1.5mu t = \frac 23 (t)^{\frac32}=\frac23\tan t\sqrt{\tan t}.$$
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As you have noted, your integral simplifies to $$2\int\sqrt{\tan x}\ \sec^2x\ dx$$ If one makes the substitution $u=\tan x$, one gets $du=\sec^2x dx$, which reduces our integral to $$2\int u^{1/2}du$$ $$=2\frac{u^{3/2}}{3/2}+C$$ $$=\frac{4u^{3/2}}{3}+C$$ $$=\frac{4\tan^{3/2}x}{3}+C$$
answered Nov 3, 2018 at 20:10