Answer: \(\displaystyle \int \mathrm{e}^{5x-2}\,\mathrm{d}x=\frac15\mathrm{e}^{5x-2}+c\), \(\displaystyle \int\frac1{4-3x}\,\mathrm{d}x=-\frac13\ln(4-3x)+c\), and \(\displaystyle \int_{\pi/3}^{\pi/2}\sec^2\left(\frac12x\right)\,\mathrm{d}x=2\left(1-\frac{\sqrt3}{3}\right)\).
(a)(i) Since the derivative of \(5x-2\) is \(5\),
\(\displaystyle \int \mathrm{e}^{5x-2}\,\mathrm{d}x=\frac15\mathrm{e}^{5x-2}+c\).
(a)(ii) Let
\(u=4-3x\).
Then
\(\dfrac{\mathrm{d}u}{\mathrm{d}x}=-3\),
so
\(\mathrm{d}x=-\dfrac13\,\mathrm{d}u\).
Therefore
\(\displaystyle \int\frac1{4-3x}\,\mathrm{d}x=-\frac13\int\frac1u\,\mathrm{d}u\).
So
\(\displaystyle \int\frac1{4-3x}\,\mathrm{d}x=-\frac13\ln u+c\).
Substituting back \(u=4-3x\), and using \(x\lt\frac43\) so that \(4-3x\gt0\),
\(\displaystyle \int\frac1{4-3x}\,\mathrm{d}x=-\frac13\ln(4-3x)+c\).
(b) Since
\(\dfrac{\mathrm{d}}{\mathrm{d}x}\tan\left(\frac12x\right)=\frac12\sec^2\left(\frac12x\right)\),
we have
\(\displaystyle \int\sec^2\left(\frac12x\right)\,\mathrm{d}x=2\tan\left(\frac12x\right)\).
Hence
\(\displaystyle \int_{\pi/3}^{\pi/2}\sec^2\left(\frac12x\right)\,\mathrm{d}x=\left[2\tan\left(\frac12x\right)\right]_{\pi/3}^{\pi/2}\).
Substitute the limits:
\(2\tan\left(\frac{\pi}{4}\right)-2\tan\left(\frac{\pi}{6}\right)\).
Since \(\tan\frac{\pi}{4}=1\) and \(\tan\frac{\pi}{6}=\frac1{\sqrt3}=\frac{\sqrt3}{3}\), the integral is
\(2(1)-2\left(\frac{\sqrt3}{3}\right)=2\left(1-\frac{\sqrt3}{3}\right)\),
as required.
