How do you integrate #int (tan2x+cot2x)^2# using substitution?

Question

Christopher Allers · Accepted Answer

#=1/2tan 2x-1/2cot2x+C#

There is no need to use substitution, it makes the integral more difficult. Simple trig identities will reduce to it to something more manageable

#int((tan2x+cot2x)^2)dx#

multiply out

#int(tan^2 2x+2tan2xcot2x+cot^2 2x)dx#

#tan2x=1/(cot2x)=>tan2xcot2x=1#

#=int(tan^2 2x+2+cot^2 2x)dx#

now #1+tan^2 theta=sec^2 theta#

and #1+cot^2 theta =csc^2 theta#

so the integral becomes

#int(sec^ 2 2x +csc^2 2x)dx#

these are standard integrals

so we have

#=1/2tan 2x-1/2cot2x+C#

Paisley Johnson · Answer

If you wanted to use substitution this is a possible solution

we have to rearrange the integral first

#int(tan 2x+cot2x)^2dx=int (tan2x+1/(tan2x))^2dx#

#=int ((tan^2 2x+1)/(tan2x))^2dx=int((tan^2 2x+1)(tan^2 2x+1))/(tan^2 2x)dx#

#=int(sec^2 2x)(tan^2 2x+1)/(tan^2 2x)dx=#

now substitute

#u=tan2x=>du=2sec^ 2xdx#

#=intcancel((sec^2 2x))(u^2+1)/u^2xx(du)/(2cancel((sec^2 2x))#

#=1/2int(1+1/u^2)dx#

#=1/2(u-1/u)+C#

#=1/2(tan2x-1/tan2x)+C#

#=1/2(tan2x-cot2x)+C#

Christopher Allers · Answer

undefined To integrate $\int (\tan^2(2x) + \cot^2(2x))^2 \, dx$ using substitution, let $u = \tan(2x)$. Then, $du = 2\sec^2(2x) \, dx$ or $dx = \frac{1}{2\sec^2(2x)} \, du$. Substituting these into the integral, we have:

$$
\begin{aligned}
\int (\tan^2(2x) + \cot^2(2x))^2 \, dx &= \int \left(u^2 + \frac{1}{u^2}\right)^2 \frac{1}{2\sec^2(2x)} \, du \
&= \frac{1}{2} \int \left(u^2 + \frac{1}{u^2}\right)^2 \sec^2(2x) \, du
\end{aligned}
$$

Using the identity $\sec^2(2x) = 1 + \tan^2(2x)$, we can rewrite the integral as:

$$
\frac{1}{2} \int \left(u^2 + \frac{1}{u^2}\right)^2 (1 + u^2) \, du
$$

Expanding and integrating term by term yields the result.