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What is integral of #tan^2(x) sec^4(x) dx?#

Answer 1

Another way of seeing this problem

#I = inttan^2(x)sec^4(x)dx# # = inttan^2(x)(1+tan^2(x))^2dx# #u = tan(x)# then #dx = (du)/(1+u^2)# #=int(u^2(1+u^2)^2/(1+u^2))du# #=int(u^2+u^4)du# #=u^3/3+u^5/5+C# #I=tan^3(x)/3+tan^5(x)/5+C#

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Answer 2

Adam Adkins

To integrate ( \tan^2(x) \sec^4(x) , dx ), you can use the identity ( \sec^2(x) = 1 + \tan^2(x) ) and express the integrand in terms of ( \tan(x) ) and its derivatives:

Given:

[ \int \tan^2(x) \sec^4(x) , dx ]

Since ( \sec^4(x) = (\sec^2(x))^2 = (1 + \tan^2(x))^2 ), the integral becomes:

[ \int \tan^2(x) (1 + \tan^2(x))^2 , dx ]

Let ( u = \tan(x) ), hence ( du = \sec^2(x) , dx ).

The integral in terms of ( u ) is:

[ \int u^2 (1 + u^2)^2 , du ]

Expand and integrate term by term:

[ \int (u^2 + 2u^4 + u^6) , du = \frac{u^3}{3} + \frac{2u^5}{5} + \frac{u^7}{7} + C ]

Substitute back ( u = \tan(x) ):

[ \frac{\tan^3(x)}{3} + \frac{2\tan^5(x)}{5} + \frac{\tan^7(x)}{7} + C ]

This is the integrated result of ( \int \tan^2(x) \sec^4(x) , 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.