Answer: (a)(i) \(-\dfrac1{50(10x-1)^5}+C\); (a)(ii) \(\dfrac23x^6+\dfrac{20}{3}x^3+25\ln x+C\); (b)(i) \(3\operatorname{sec}^2(3x+1)\); (b)(ii) \(0.3222\) to 4 significant figures.
(a)(i) Write the integrand as
\((10x-1)^{-6}.\)
Using the reverse chain rule,
\(\int(10x-1)^{-6}\,dx =-\frac1{50}(10x-1)^{-5}+C.\)
So
\(\int\frac1{(10x-1)^6}\,dx =-\frac1{50(10x-1)^5}+C.\)
(a)(ii) First expand the numerator:
\((2x^3+5)^2=4x^6+20x^3+25.\)
Then
\(\frac{(2x^3+5)^2}{x}=4x^5+20x^2+\frac{25}{x}.\)
Integrating term by term gives
\(\int\frac{(2x^3+5)^2}{x}\,dx =\frac23x^6+\frac{20}{3}x^3+25\ln x+C.\)
(b)(i) Since \(\dfrac{d}{dx}\tan u=\operatorname{sec}^2u\,\dfrac{du}{dx}\),
\(\frac{dy}{dx}=3\operatorname{sec}^2(3x+1).\)
(b)(ii) From part (b)(i),
\(\int\frac{\operatorname{sec}^2(3x+1)}2\,dx =\frac16\tan(3x+1).\)
Also,
\(\int-\sin x\,dx=\cos x.\)
Therefore
\(\int_{\pi/12}^{\pi/10}\left(\frac{\operatorname{sec}^2(3x+1)}2-\sin x\right)\,dx =\left[\frac16\tan(3x+1)+\cos x\right]_{\pi/12}^{\pi/10}.\)
Evaluating,
\(\left(\frac16\tan\left(\frac{3\pi}{10}+1\right)+\cos\frac{\pi}{10}\right) -\left(\frac16\tan\left(\frac{\pi}{4}+1\right)+\cos\frac{\pi}{12}\right).\)
This gives
\(0.3222\)
to 4 significant figures.
