Answer: (a)(ii) \(3\operatorname{sec}x-\frac43e^{3x/4}+c\). (b) \(p=10\).

(a)(i) Since

\(\mathrm{f}(x)=\frac1{\cos x}=(\cos x)^{-1},\)

differentiate using the chain rule:

\(\mathrm{f}'(x)=-(\cos x)^{-2}(-\sin x)=\frac{\sin x}{\cos^2x}.\)

But

\(\frac{\sin x}{\cos^2x}=\frac{\sin x}{\cos x}\cdot\frac1{\cos x}=\tan x\operatorname{sec}x.\)

Therefore

\(\mathrm{f}'(x)=\tan x\operatorname{sec}x.\)

(a)(ii) From part (i),

\(\int \tan x\operatorname{sec}x\,dx=\operatorname{sec}x.\)

Also

\(\sqrt[4]{e^{3x}}=e^{3x/4}.\)

Hence

\(\int\left(3\tan x\operatorname{sec}x-\sqrt[4]{e^{3x}}\right)\,dx =3\operatorname{sec}x-\int e^{3x/4}\,dx.\)

Now

\(\int e^{3x/4}\,dx=\frac43e^{3x/4}.\)

Therefore

\(\boxed{3\operatorname{sec}x-\frac43e^{3x/4}+c}.\)

(b)

\(\int\frac{p}{px+10}\,dx=\ln(px+10)+c.\)

So

\(\int_2^5\frac{p}{px+10}\,dx=\ln(5p+10)-\ln(2p+10).\)

Given that this is \(\ln2\),

\(\ln\left(\frac{5p+10}{2p+10}\right)=\ln2.\)

Thus

\(\frac{5p+10}{2p+10}=2.\)

Therefore

\(5p+10=4p+20,\)

so

\(\boxed{p=10}.\)