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How do you evaluate the integral #int xsec(4x^2+7)#?

Answer 1

Isaiah Allen

# ln|sec(4x^2+7)+tan(4x^2+7)|+C.#

If we subst. #4x^2+7=y," then, "8xdx=dy.#

Hence,

#intxsec(4x^2+7)dx=1/8int(sec(4x^2+7))8xdx,#

#=1/8intsecydy,#

#=ln|secy+tany|,#

#=ln|sec(4x^2+7)+tan(4x^2+7)|+C.#

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Answer 2

Elijah Abram

The answer is #=1/8ln(tan(4x^2+7)+sec(4x^2+7))+C#

We need

#intsecxdx=ln(tanx+secx)+C#

We perform this integral by substitution

Let

#u=4x^2+7#, #=>#, #du=(8x)dx#

Therefore,

#intxsec(4x^2+7)dx=1/8intsecudu#

#=1/8ln(tanu + secu)#

#=1/8ln(tan(4x^2+7)+sec(4x^2+7))+C#

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Answer 3

Luke Alvarez

To evaluate the integral ∫xsec(4x^2+7), we can use u-substitution. Let u = 4x^2 + 7. Then du/dx = 8x, which implies dx = du/(8x).

Substituting u and dx into the integral, we get:

∫xsec(u) * (1/(8x)) du

Simplify to get:

(1/8) * ∫sec(u) du

Now, integrate sec(u) with respect to u:

(1/8) * ln|sec(u) + tan(u)| + C

Finally, substitute back u = 4x^2 + 7:

(1/8) * ln|sec(4x^2 + 7) + tan(4x^2 + 7)| + 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.