Find the antiderivative/integral of (tan 2x + cot 2x)^2 dx. Show step by step solution.?

Answer 1

Integral ( [tan(2x) + cot(2x)]^2 dx )

First, expand the square binomial.

Integral ( [tan^2(2x) + 2tan(2x)cot(2x) + cot^2(2x)] dx )

Note that tan(2x) and cot(2x) are reciprocals of each other (one is sine over cos, the other is cos over sine), so they should cancel into 1.

Integral ( [tan^2(2x) + 2 + cot^2(2x)] dx )

Use these identities:

tan^2(y) = sec^2(y) - 1
cot^2(y) = csc^2(y) - 1

In our case, y = 2x, so we have

Integral [ sec^2(2x) - 1 + 2 + csc^2(2x) - 1 ] dx

Notice the -1, the -1 and the 2 will just cancel out.

Integral [ sec^2(2x) + csc^2(2x) ] dx

These two are *almost* known derivatives, with one difference being a 2x in the inside, and another difference being that
d/dx cot(x) = -csc^2(x). Note that if you differentiate tan(2x), you would get 2sec^2(2x) due to the chain rule, so when integrating, just offset it by (1/2). When differentiating cot(2x), you would obtain -2csc^2(2x), so we offset the integral by (-1/2). Therefore, our answer is

(1/2)tan(2x) + (-1/2)cot(2x) + C

Simplifying,

(1/2)tan(2x) - (1/2)cot(2x) + C

Answer 2

Integrate Tan 2x

Answer 3

the truthful answer is
simplify the tan 2x + cot 2x first you do that by changing it to
(sin(2x)/cos(2x))+(cos(2x)/sin(2x))
this becomes
(sin^2(2x)+cos^2(2x))/(sin(2x)cos(2x))
sin^2(2x)+cos^2(2x) = 1 so
1/(sin(2x)cos(2x))
you then notice that sin(2x)cos(2x) is close to 2sin(2x)cos(2x) which = sin(4x) so you times both the top and bottom by 2 and get
2/(2sin(2x)cos(2x)) so
2/(sin(4x)) = 2csc(4x)
now remember we just simplified the part the is to be squared, so now square it and you get
4csc^2(4x)
now find the integral
4csc^2(4x)dx
set u=4x and du=4dx
csc^2(u)du
-cot(u)+c and plug u back in
-cot(4x)+c

I checked this online at http://integrals.wolfram.com/index.jsp and made sure it is the correct answer. Feel free to use it to check solutions.

Answer 4

∫(tan (2x) + cot (2x)² dx
∫tan² (2x) + 2 + cot² (2x) dx
∫(tan² (2x) + 1) + (cot² (2x) + 1) dx
∫sec² (2x) + csc² (2x) dx
tan (2x)/2 - cot (2x)/2 + C

Answer 5

I am in full agreement with and fully endorse what Mr."Aryan" has stated in his answer which is not only very detailed but very precise ,to the point and very crisp too.

Answer 6

Do your own homework.