Question Number 79964 by ubaydulla last updated on 29/Jan/20
![∫(tan^2 x+tan^4 x)dx=[tan^2 x=t]=∫(t^2 +t^4 )dx=(t^3 /3)+(t^5 /5)+c](Q79964.png)
$$\int\left(\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{tan}\:^{\mathrm{4}} {x}\right){dx}=\left[\mathrm{tan}\:^{\mathrm{2}} {x}={t}\right]=\int\left({t}^{\mathrm{2}} +{t}^{\mathrm{4}} \right){dx}=\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+\frac{{t}^{\mathrm{5}} }{\mathrm{5}}+{c} \\ $$
Commented by $@ty@m123 last updated on 29/Jan/20
$${You}\:{cannot}\:{integrate}\:{function}\:{of} \\ $$$${t}\:{with}\:{respect}\:{to}\:{x}. \\ $$$${If}\:{t}=\mathrm{tan}\:{x} \\ $$$${dt}=\mathrm{sec}\:^{\mathrm{2}} {xdx} \\ $$$${Now}\:{try}\:{again}. \\ $$
Commented by john santu last updated on 29/Jan/20
$$\int\:\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}\right)\mathrm{dx}= \\ $$$$\int\mathrm{tan}\:^{\mathrm{2}} \mathrm{xsec}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}\:= \\ $$$$\int\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{d}\left(\mathrm{tan}\:\mathrm{x}\right)= \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}\:+\mathrm{c} \\ $$
Commented by key of knowledge last updated on 29/Jan/20
$$\int\mathrm{t}^{\mathrm{n}} .\mathrm{dt}=\frac{\mathrm{t}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}\:\:\:\:\:\int\mathrm{t}^{\mathrm{n}} .\mathrm{dx}\neq\frac{\mathrm{t}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}} \\ $$