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Question Number 79964 by ubaydulla last updated on 29/Jan/20

$$\int (\tan {}^{2}x+\tan {}^{4}x)dx=[\tan {}^{2}x=t]=\int (t^2+t^4)dx=\frac{t^3}{3}+\frac{t^5}{5}+c$$

Commented by $@ty@m123 last updated on 29/Jan/20

$$Youcannotintegratefunctionof$$$$twithrespecttox.$$$$Ift=\tan x$$$$dt=\sec {}^{2}xdx$$$$Nowtryagain.$$

Commented by john santu last updated on 29/Jan/20

$$\int \tan {}^2\mathrm{x}(1+\tan {}^2\mathrm{x})\mathrm{dx}=$$$$\int \tan {}^2\mathrm{xsec}{}^2\mathrm{x}\mathrm{dx}=$$$$\int \tan {}^2\mathrm{x}\mathrm{d}\left(\tan \mathrm{x}\right)=$$$$\frac{1}{3}\tan {}^3\mathrm{x}+\mathrm{c}$$

Commented by key of knowledge last updated on 29/Jan/20

$$\int {\mathrm{t}}^{\mathrm{n}}.\mathrm{dt}=\frac{{\mathrm{t}}^{\mathrm{n}+1}}{\mathrm{n}+1}\int {\mathrm{t}}^{\mathrm{n}}.\mathrm{dx}\ne \frac{{\mathrm{t}}^{\mathrm{n}+1}}{\mathrm{n}+1}$$

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