Solve ∫(tanxsec^{2}x)dx Integral with Substitutions

[t6x3]

∫(tanxsec$^{2}x$)dx <---Original function to integrate.

I do:

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


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

 

[t6x3]

Hey guys, searching around the web I found an explanation from MIT for this very same integral, here it is:

ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-c-mean-value-theorem-antiderivatives-and-differential-equations/session-38-integration-by-substitution/MIT18_01SCF10_ex38sol.pdf

So we know both my answer and the book's are correct but I still want to know how these differences among answers would affect the computation of a definite integral?

 

[Karamata]

Your result is good. They were wrong.

 

[theorem4.5.9]

Actually, you and the text are right!

$\frac{1}{2}\mathrm{tan}^2(\theta) = \frac{1}{2}\frac{\mathrm{sin}^2(\theta)}{\mathrm{cos}^2(\theta )} = \frac{1}{2}\frac{1 - \mathrm{cos}^2(\theta )}{\mathrm{cos}^2(\theta)} = \frac{1}{2}\frac{1}{\mathrm{cos}^2(\theta)}-\frac{1}{2}\frac{\mathrm{cos}^2(\theta)}{\mathrm{cos}^2(\theta)}$

So just absorb the $-\frac{1}{2}$ into the constant.

To answer your question about definite integral: As you know the arbitrary constants will cancel out. The above computation shows that both indefinite integrals yield the same definite integral (regardless of the constant you pick).

 

[Karamata]

So just absorb the $-\frac{1}{2}$ into the constant.

Hehe, interesting. Thanks.