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
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.
Answers & Comments
Verified answer
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
Integrate Tan 2x
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.
∫(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
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.
Do your own homework.