Evaluate the following integrals:

We know $d(\cos x)=\sin x$, and $\tan$ can be written interms of $\cos$ and $\sin$

$\therefore \tan x=\frac{\sin x}{\cos x}$

$\therefore$ The given equation can be written as

$\Rightarrow \int \frac{\sin x}{\cos x \sqrt{\cos x}} d x$

$\Rightarrow \int \frac{\sin x}{\cos ^{3} \backslash^{2} x} d x$

Now assume $\cos x=t$

$d(\cos x)=-d t$

$\sin x d x=-d t$

Substitute values of $\mathrm{t}$ and $\mathrm{dt}$ in above equation

$\Rightarrow \int \frac{-\mathrm{dt}}{\mathrm{t}^{3} / 2}$

$\Rightarrow-\int t^{-3 \backslash 2} d t$

$\Rightarrow 2 t^{-1 \backslash 2}+c$

$\Rightarrow 2 \cos ^{-1 \backslash 2} x+c$

$\Rightarrow \frac{2}{\sqrt{\cos x}}+C$