∫(tanxsec[itex]^{2}x[/itex])dx <---Original function to integrate.(adsbygoogle = window.adsbygoogle || []).push({});

I do:

∫(tanxsec[itex]^{2}x[/itex])dx ---> ∫(tanx[itex]\frac{1}{cos^{2}x}[/itex])dx ---> ∫(([itex]\frac{sinx}{cosx}[/itex])([itex]\frac{1}{cos^{2}x}[/itex]))dx ---> ∫([itex]\frac{sinx}{cos^{3}x}[/itex])dx ---> substitution: [t=cosx , dt=-sinxdx -> [itex]\frac{dt}{-sinx}[/itex]=dx] ---> ∫([itex]\frac{sinx}{t^{3}}[/itex])[itex]\frac{dt}{-sinx}[/itex] ---> -∫(t[itex]^{-3}[/itex])dt ---> -[itex]\frac{t^{-2}}{-2}[/itex] + k ---> [itex]\frac{1}{2t^{2}}[/itex] + k ---> [itex]\frac{1}{2cos^{2}x}[/itex] + k <---My answer

However my book gives [itex]\frac{tan^{2}x}{2}[/itex] +k as an answer, I followed their process and understand what they did, I just want to know if my answer is also correct andwhat would happend if this was a definite integral instead of an indefinite one? would arriving at different results create problems when computing definite integrals?

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# Indefinite integral. Different results depending upon what substitutions you use.

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