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How do you integrate #int xtan(x^2)sec(x^2)# using substitution?

Answer 1

#1/2sec(x^2)+C#

#I=intxtan(x^2)sec(x^2)dx#

The first substitution we will make is #u=x^2#. Differentiating both sides gives us #(du)/dx=2x#, so #du=2xdx#. Notice that in our integrand we currently have #xdx#, so multiply the interior by #2# and the exterior by #1/2# to balance ourselves out.

#I=1/2int2xtan(x^2)sec(x^2)dx#

#I=1/2inttan(x^2)sec(x^2)(2xdx)#

Substituting in our values for #u# and #du#:

#I=1/2inttan(u)sec(u)du#

This is the integral for #sec(u)#, since #d/(du)sec(u)=sec(u)tan(u)#.

#I=1/2sec(u)+C#

Since #u=x^2#:

#I=1/2sec(x^2)+C#

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Answer 2

William Abdalla

To integrate (\int x\tan(x^2)\sec(x^2)) using substitution, let (u = x^2). Then, (du = 2x dx) or (x dx = \frac{1}{2} du).

The integral becomes: [\int x\tan(x^2)\sec(x^2) dx = \frac{1}{2} \int \tan(u) \sec(u) du]

Now, recall that (\tan(u)\sec(u) = \sec(u)\tan(u)). So, the integral becomes: [\frac{1}{2} \int \sec(u) \tan(u) du = \frac{1}{2} \sec(u) + C]

Finally, substitute back (u = x^2) to get the final result: [\frac{1}{2} \sec(x^2) + C]

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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.