Question

How do I properly integrate sec^4 (x)?

Answer 1

\[\int\limits \sec^4 (x) dx = (\sec^2 (x))^2 dx\]
and
\[\sec^2 (x) = 1+\tan^2 (x)\]
so
\[\int\limits\limits (1+\tan^2 (x)) \sec^2 (x) dx\]

Answer 2

u = tan(x) and du = sec^2 (x)dx

Answer 3

so \[\int\limits (1+u^2) du = u +\frac{ 1 }{ 3 }u^3 + C\]
equals
\[\tan(x) + \frac{ 1 }{ 3 }\tan^3 (x) + C\]

Is this a good way to approach the problem?

Answer 4

\[\int\limits \sec ^4(x) \, dx=\frac{2 \tan (x)}{3}+\frac{1}{3} \tan (x) \sec ^2(x) \]

Answer 5

in your second comment you went off
you should have \[\int (1+\tan^2x)^2dx\] what you did was wrong don't where did you brought that sec^2x there

Answer 6

this is what you did actually
\[\int \sec^4xdx=\int \sec^2x\sec^2xdx=\int (1+\tan^2x)\sec^2xdx\]
then do u-sub
u=tanx then du=sec^2xdx

Answer 7

your result seems fine to me