 Pretoto4o

2021-12-05

Evaluate each integral.
${\int }_{0}^{4}\frac{{\mathrm{sec}h}^{2}\sqrt{x}}{\sqrt{x}}dx$ Opeance1951

Step 1
We have to find definite integral:
${\int }_{0}^{4}\frac{{\mathrm{sec}h}^{2}\sqrt{x}}{\sqrt{x}}dx$
This will be solved by substitution method or change of variable.
Assuming $t=\sqrt{x}$
Limit will change as following,
$x⇒0⇒t⇒0$
$x⇒4⇒t⇒2$
Differentiating with respect to x the assumed value, we get
$\frac{dt}{dx}=\frac{d\sqrt{x}}{dx}$
$\frac{dt}{dx}=\frac{1}{2\sqrt{x}}$
$2dt=\frac{dx}{\sqrt{x}}$
Step 2
Substituting above value in the definite integral, we get
${\int }_{0}^{4}\frac{{\mathrm{sec}h}^{2}\sqrt{x}}{\sqrt{x}}dx={\int }_{0}^{2}{\mathrm{sec}h}^{2}t\left(2dt\right)$
$=2{\int }_{0}^{2}\mathrm{sec}{h}^{2}tdt$
$=2{\left[\mathrm{tan}ht\right]}_{0}^{2}$ (since $\int {\mathrm{sec}h}^{2}xdx=\mathrm{tan}hx+C$)
$=2\left[\mathrm{tan}h2-\mathrm{tan}h0\right]$
$=2\left[\mathrm{tan}h2-0\right]$
$=2\mathrm{tan}h2$
Hence, value of definite integral is $2\mathrm{tan}h\left(2\right)$.

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