How do you integrate int tan^5(2x)sec^2(2x)?

2 Answers

\frac{\tan^6(2x)}{12}+C

Explanation:

Let \tan (2x)=u\implies 2\sec^2(2x)\ dx=du

\therefore \int \tan^5(2x)\sec^2(2x)\ dx

=\int u^5\ {du}/{2}

=1/2\int u^5\ du

=1/2 \cdot u^6/6+C

=\frac{\tan^6(2x)}{12}+C

Jul 27, 2018

tan^6(2x)/12+c

Explanation:

We'll have to make a trigonometric substitution here, and we have tangents and secants, so we're looking for either a tangent with an odd power or a secant with an event power.

Since we have both in this case, it does not matter which one we choose for substitution. In this particular case, it seems like using tan(2x) would be easier because we already have a sec^2(2x).

So:

color(red)(u = tan(2x)
color(blue)(du = 2sec^2(2x)dx

Plugging that back into the original integral:

inttan^5(2x)sec^2(2x)dx

int2/2*tan^5(2x)sec^2(2x)dx

int1/2*color(red)(tan)^5color(red)((2x))*color(blue)(2sec^2(2x)dx

int1/2*color(red)(u)^5color(blue)(du

1/12u^6+c

Substitute u back in:

1/12tan^6(2x)+c or tan^6(2x)/12+c