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How do you find the integral of #x^2(sec(x^3))^2#?

Answer 1

Emilia Aguilar

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

We want to integrate:

#intx^2(sec(x^3))^2dx#

Let #u=x^3=>du=3x^2dx#. Then, multiply the integrand by #3# and the exterior of the integral expression by #1/3#.

#1/3int3x^2(sec(x^3))^2dx#

Substitute in #u# and #du#.

#=1/3int(sec(u))^2du#

Note that the derivative of #tan(u)# is #(sec(u))^2#, so the integral of #(sec(u))^2# is #tan(u)+C#.

#=1/3tan(u)+C#

Since #u=x^3#:

#=1/3tan(x^3)+C#

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Answer 2

Emilia Aguilar

To find the integral of (x^2(\sec(x^3))^2), you can use the substitution method. Let (u = x^3), then (du = 3x^2 dx). Rearranging, (dx = \frac{du}{3x^2}). Now, substitute these into the integral:

[\int x^2(\sec(x^3))^2 , dx = \frac{1}{3} \int \frac{u^2}{\sec^2(u)} , du]

Since (\sec^2(u) = 1 + \tan^2(u)), we have:

[\frac{1}{3} \int \frac{u^2}{1 + \tan^2(u)} , du]

Now, you can use the trigonometric identity (\tan^2(u) + 1 = \sec^2(u)) to simplify:

[\frac{1}{3} \int \frac{u^2}{\sec^2(u)} , du]

[= \frac{1}{3} \int u^2 , du]

[= \frac{1}{3} \left(\frac{u^3}{3} + C\right)]

Finally, substitute back (u = x^3):

[= \frac{1}{9}x^9 + 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.