integral

[math]

how would you integrate tan^4 (x) / sec^2 (x) dx
thanks

 

[pka]

These will help you.
\(\displaystyle \L\begin{array}{l}
\tan ^2 (x) = \sec ^2 (x) - 1 \\
\tan ^4 (x) = \sec ^4 (x) - 2\sec ^2 (x) + 1 \\
\frac{{\tan ^4 (x)}}{{\sec ^2 (x)}} = \sec ^2 (x) - 2 + \cos ^2 (x) \\
\end{array}\)

 

[math]

Ok, so I know cos2x = 2cos^2(x) -1
cos^2 (x) = 1/2 + cos(2x) / 2

tan^4(x) / sec^2(x) = sec^2(x) - 2 + cos^2(x)

integral of (1 / (1/2 + cos(2x)/2)) - 2 + 1/2 + cos(2x)/2 dx

I don't know how to find the integral of this part: (1 / (1/2 + cos(2x)/2)) which is sec^2(x)


thanks

 

[pka]

Don't complicate things!
\(\displaystyle \L\int {\sec ^2 (x)dx = \tan (x)}\)

 

[math]

oops! thanks!

so, does the rest of this look right?

tanx - 2x + x/2 + sin(2x)/4
tanx - 3x/2 + sin(2x)/4 + c

 

[morson]

Looks fine.