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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 :)

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