Question

Find the indefinite integral. $$ \int \sec t(\sec t+\tan t) d t $$

Short answer

The indefinite integral of \(\int \sec t(\sec t+\tan t) d t\) is \(\frac{1}{3} (\sec t)^3 + \frac{1}{2} (\sec t)^2 + C\)

Step-by-step solution

Identify a possible substitution

Noticing that the derivative of \(\sec t\) is \(\sec t \tan t\), it might simplify things to let \(u = \sec t\). This would mean our differential \(du\) would then be \(\sec t \tan t dt\).

Rewrite the integral in terms of \(u\)

Substituting \(u = \sec t\) into the integral, this then becomes \(\int u (u+ du)\). This simplifies down to \(\int (u^2 du+u du)\), leading to \(\int u^2 du + \int u du\).

Evaluate the integrals

Integrating these two separate integrals results in the expressions \(\frac{1}{3} u^3 + \frac{1}{2} u^2\).

Substitute back for \(t\)

Finally, replace the \(u\) back into the equation with \(\sec t\), resulting in the final answer \(\frac{1}{3} (\sec t)^3 + \frac{1}{2} (\sec t)^2 + C\)