Question

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

Answer 1

`\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`

Answer 2

`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`