How do you find #int cot^2x*tan^2xdx#?

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Answer 1 · 2018-04-10T19:16:38

#I=x+c#

We know that,

#color(red)((1)tantheta*cottheta=1#

Here,

#I=intcot^2xtan^2xdx#

#I=int(cotxtanx)^2dx...toApply(1)#

#=int(1)^2dx#

#=x+c#

Note:The question is more related to trigonometry than calculus ?!!!

Answer 2 · 2018-04-10T19:16:55

#x+c#

#rarr#where #c# is the constant of integration.

Note that : #cot^2x=1/tan^2x#
#rArr cot^2x.tan^2x =(1/cancel(tan^2x)).cancel(tan^2x)=1#

Now we have only left #1.dx# so our integral becomes :-

#intcot^2x.tan^2xdx=int1dx#

#:.intcot^2x.tan^2xdx=x+c# ; for some constant #c#