Mathematics
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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