Home > Calculus > Integration by Trigonometric Substitution

How do you integrate #sec2x tan2x dx#?

Answer 1

Isaiah Allen

I would write the integral as:

#intsec(2x)tan(2x)dx=int1/cos(2x)sin(2x)/cos(2x)dx=#

we can see that:

#d[cos(2x)]=-2sin(2x)dx# and so: #sin(2x)dx=(d[cos(2x)])/-2#

with this in mind I write the integral as:

#int1/cos^2(2x)*1/-2d[cos(2x)]=# #=1/-2intcos^(-2)(2x)d[cos(2x)]=# (treating #cos(2x)# as if it were #x# in a normal integration where you get #x^(n+1)/(n+1)#) #=1/2*1/cos(2x)+c#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Emma Aldridge

To integrate sec^2(x) tan^2(x) dx, you can use the substitution method.

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

The integral becomes:

∫ tan^2(x) sec^2(x) dx = ∫ u^2 du.

Now integrate u^2 with respect to u:

∫ u^2 du = (1/3)u^3 + C.

Substitute back u = tan(x):

(1/3)tan^3(x) + C.

By signing up, you agree to our Terms of Service and Privacy Policy

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.