Question

How to solve these integrals?
integrate (secx)^2 tanx dx
integrate xsec(x^2)tan(x^2) dx

Answer 1

For the first one:\(u=\tan x,\ du=\sec^2x\ dx\).

Answer 2

For the second one: \(u=x^2,\ du=2x\ dx\), so \[
\int x\sec(x^2)\tan(x^2)dx=\frac{1}{2}\int\sec u\tan u\ du=\frac{1}{2}\sec u.
\]

Answer 3

for the first one using u=tanx, I get an answer of 0.5(tanx)^2+C when the actual answer is 0.5(secx)^2+C (by using u=secx) shouldn't both substitutions give the same answer.

Answer 4

Those two solutions are actually equivalent. Remember, \(\tan^2x+1=\sec^2x\), so \(\tan^2x+1+C=\sec^2x+C\), and \(C+1=C\).

Answer 5

For further proof, derive both \(\tan^2x\) and \(\sec^2x\). You'll get the same thing :)