tan-2-x-tan-4-x-dx-tan-2-x-t-t-2-t-4-dx-t-3-3-t-5-5-c-

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

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

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

You cannot integrate function of t with respect to x. If t=tan x dt=sec^2 xdx Now try again.

$${You}\:{cannot}\:{integrate}\:{function}\:{of} \\ $$$${t}\:{with}\:{respect}\:{to}\:{x}. \\ $$$${If}\:{t}=\mathrm{tan}\:{x} \\ $$$${dt}=\mathrm{sec}\:^{\mathrm{2}} {xdx} \\ $$$${Now}\:{try}\:{again}. \\ $$

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

∫ tan^2 x(1+tan^2 x)dx= ∫tan^2 xsec^2 x dx = ∫tan^2 x d(tan x)= (1/3)tan^3 x +c

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

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

∫t^n .dt=(t^(n+1) /(n+1)) ∫t^n .dx≠(t^(n+1) /(n+1))

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

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tan-2-x-tan-4-x-dx-tan-2-x-t-t-2-t-4-dx-t-3-3-t-5-5-c-

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