Harold Kessler

2021-12-30

Evaluate the indefinite integral.
$\int \frac{{\mathrm{sec}}^{2}\left(\sqrt{x}\right)dx}{\sqrt{x}}$

vicki331g8

Expert

Step 1
We have to evaluate the integral:
$\int \frac{{\mathrm{sec}}^{2}\left(\sqrt{x}\right)dx}{\sqrt{x}}$
We will use substitution method since derivatives of one function is present in the integral.
So assuming,
$t=\sqrt{x}$
differentiating with respect to x,
$\frac{dt}{dx}=\frac{d\sqrt{x}}{dx}$
$\frac{dt}{dx}=\frac{1}{2\sqrt{x}}$
$2dt=\frac{dx}{\sqrt{x}}$
Substituting above values, we get
$\int \frac{{\mathrm{sec}}^{2}\left(\sqrt{x}\right)dx}{\sqrt{x}}=\int {\mathrm{sec}}^{2}\left(t\right)\left(2dt\right)$
$=2\int {\mathrm{sec}}^{2}\left(t\right)dt$
Step 2
Since we know that
$\int {\mathrm{sec}}^{2}xdx=\mathrm{tan}x+C$
therefore,
$2\int {\mathrm{sec}}^{2}\left(t\right)dt=2\mathrm{tan}\left(t\right)+C$
$=2\mathrm{tan}\left(\sqrt{x}\right)+C$
Where, C is an arbitrary constant.
Hence, value of given integral is $2\mathrm{tan}\left(\sqrt{x}\right)+C$.

usumbiix

Expert

$\int \frac{{\mathrm{sec}}^{2}\left(\sqrt{x}\right)}{\sqrt{x}}dx$
Substitution $u=\sqrt{x}⇒\frac{du}{dx}=\frac{1}{2\sqrt{x}}⇒dx=2\sqrt{x}du$
$=2\int {\mathrm{sec}}^{2}\left(u\right)du$
$\int {\mathrm{sec}}^{2}\left(u\right)du$
This is the well-known tabular integral:
$=\mathrm{tan}\left(u\right)$
We substitute the already calculated integrals:
$2\int {\mathrm{sec}}^{2}\left(u\right)du$
$=2\mathrm{tan}\left(u\right)$
Reverse replacement $u=\sqrt{x}:$
$=2\mathrm{tan}\left(\sqrt{x}\right)$
Solution:
$=2\mathrm{tan}\left(\sqrt{x}\right)+C$

karton

Expert

$\int \frac{\mathrm{sec}\left(\sqrt{x}{\right)}^{2}}{\sqrt{x}}dx$
$\int 2\mathrm{sec}\left(t{\right)}^{2}dt$
$2×\int \mathrm{sec}\left(t{\right)}^{2}dt$
$2\mathrm{tan}\left(t\right)$
$2\mathrm{tan}\left(\sqrt{x}\right)$
Add $C\in \mathbb{R}$
Solution:
$2\mathrm{tan}\left(\sqrt{x}\right)+C,C\in \mathbb{R}$

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