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

Answer 1

Elijah Abram

#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#

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Answer 2

Emily Ammerman

#I=x+c#

As we are aware,

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

Here,

#I=intcot^2xtan^2xdx#

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

#=int(1)^2dx#

#=x+c#

It should be noted that the question is more about trigonometry than calculus.

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Answer 3

Isaiah Allen

To find ∫ cot²(x) * tan²(x) dx, use the trigonometric identity cot²(x) = csc²(x) - 1 and tan²(x) = sec²(x) - 1 to rewrite the integral. Then integrate term by term using basic trigonometric integrals. The result will be ∫ (csc²(x) - 1) * (sec²(x) - 1) dx.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.