Question

another integral problem....

Answer 1

\[\int\limits_{ }^{ }\sec^4(x) \tan^4(x)~dx\]

Answer 2

hint, please?

Answer 3

\[\int\limits_{ }^{ }\sec^4(x) \tan^4(x)~dx\]\[\large \int\limits_{ }^{ }\tan^4x \sec^2x \sec^2x~dx\]\[\large \int\limits_{ }^{ }\tan^4x (\tan^2x+1) \sec^2x~dx\]\[u=\tan(x),~~~~~du=\sec^2x~dx\]

Answer 4

\[\large \int\limits_{ }^{ }u^4(u^2+1) ~du\]

Answer 5

\[\frac{1}{7}\tan^7x+\frac{1}{5}\tan^5x +C\]right?

Answer 6

looks good

Answer 7

I was thinking of u substitution because sec^2x is the derivative of tan(x), but didn't know how to put it so that it can be substitutable.

Answer 8

I will be beack. excuse me for sec...

Answer 9

\[\frac{d}{dx}(\frac{1}{7}\tan^7(x)+\frac{1}{5}\tan^5(x)+C) \\ \frac{1}{7}(7)\tan^6(x)\sec^2(x)+\frac{1}{5}(5)\tan^4(x)\sec^2(x)+0 \\ \tan^6(x)\sec^2(x)+\tan^4(x)\sec^2(x) \\ \sec^2(x)\tan^4(x)(\tan^2(x)+1) \\ \sec^2(x)\tan^4(x)(\sec^2(x)) \\tan^4(x)\sec^4(x)\]

Answer 10

oh, tnx for checking it while I departed.