How do you integrate #int sec^2sqrtx# by integration by parts method?

Question

How do you integrate #int sec^2sqrtxdx# by integration by parts method?

Ezekiel Alexander · Accepted Answer

#2[sqrtxtan(sqrtx)-ln|tansqrtx+secsqrtx|]+C#.

Let #sqrtx=y rArr x=y^2 rArr dx=2ydy#. #:. I=intsec^2sqrtxdx=int(sec^2y)(2y)dy=2intysec^2ydy#. The Rule of Integration by Parts (IbP) states : #" (IbP) : "intuvdy=uintvdy-int((du)/dy*intvdy)dy#. We take, #u=y, and, v=sec^2y#. #:. (du)/dy=1, and, intvdy=tany#. #:. I=2[ytany-inttanydy]# #=2[ytany-ln|tany+secy|}# Replacing #y" by "sqrtx#, we have, #I=2[sqrtxtan(sqrtx)-ln|tansqrtx+secsqrtx|]+C#. Enjoy maths.!

Evelyn Agard · Answer

undefined To integrate $\int \sec^2(\sqrt{x}) \, dx$ using integration by parts, let $u = \sqrt{x}$ and $dv = \sec^2(\sqrt{x}) \, dx$. Then, $du = \frac{1}{2\sqrt{x}} \, dx$ and $v = \tan(\sqrt{x})$.

Applying the integration by parts formula:
$$ \int u \, dv = uv - \int v \, du $$

we get:
$$ \int \sec^2(\sqrt{x}) \, dx = \sqrt{x}\tan(\sqrt{x}) - \int \tan(\sqrt{x}) \frac{1}{2\sqrt{x}} \, dx $$

Now, $\int \tan(\sqrt{x}) \frac{1}{2\sqrt{x}} \, dx$ can be integrated using substitution method. Let $t = \sqrt{x}$, then $dt = \frac{1}{2\sqrt{x}} \, dx$.

Substituting $t = \sqrt{x}$ into the integral, we have:
$$ \int \tan(\sqrt{x}) \frac{1}{2\sqrt{x}} \, dx = \int \tan(t) \, dt $$

This integral can be evaluated directly to get the final result.

So, the integration of $\int \sec^2(\sqrt{x}) \, dx$ using integration by parts involves a subsequent integral of $\tan(t) \, dt$, where $t = \sqrt{x}$.