Find \(\int {\,\frac{{{x^{\frac{3}{2}}} - 8}}{{\sqrt x - 2}}dx} \).

\(\begin{array}{1}\int {\,\frac{{{x^{\frac{3}{2}}} - 8}}{{\sqrt x - 2}}dx}
= \int {\,\frac{{({x^{\frac{1}{2}}} - 2)(x + 2{x^{\frac{1}{2}}} + 4)}}{{{x^{\frac{1}{2}}} - 2}}dx} \\ = {\int {\,(x + 2{x^{\frac{1}{2}}} + 4)dx} ^{}}\\ = \frac{1}{2}{x^2} + 2 \cdot \frac{2}{3}{x^{\frac{3}{2}}} + 4x + C\\ = {\underline{\underline {\frac{1}{2}{x^2} + \frac{4}{3}{x^{\frac{3}{2}}} + 4x + C}} ^{}}\end{array}\)

Find \(\int {\,\frac{{{{\cos }^3}\theta - \cos 3\theta }}{{\cos \theta }}d\theta } + \int {\,\frac{{{{\sin }^3}\theta + \sin 3\theta }}{{\sin \theta }}d\theta } \).

\(\begin{array}{1}\int {\,\frac{{{{\cos }^3}\theta - \cos 3\theta }}{{\cos \theta }}d\theta + \int {\,\frac{{{{\sin }^3}\theta + \sin 3\theta }}{{\sin \theta }}d\theta } } \\ = \int {\,(\frac{{{{\cos }^3}\theta - \cos 3\theta }}{{\cos \theta }} + \frac{{{{\sin }^3}\theta + \sin 3\theta }}{{\sin \theta }})\,d\theta } \\ = \int {\,\frac{{\sin \theta {{\cos }^3}\theta - \sin \theta \cos 3\theta + {{\sin }^3}\theta \cos \theta + \cos \theta \sin 3\theta }}{{\sin \theta \cos \theta }}d\theta } \\ = \int {\,\frac{{(\sin \theta {{\cos }^3}\theta + {{\sin }^3}\theta \cos \theta ) + (\cos \theta \sin 3\theta - \sin \theta \cos 3\theta )}}{{\sin \theta \cos \theta }}d\theta } \\ = \int {\,\frac{{\sin \theta \cos \theta ({{\cos }^2}\theta + {{\sin }^2}\theta ) + {\textstyle{1 \over 2}}[\sin (3\theta + \theta ) + \sin (3\theta - \theta )] - {\textstyle{1 \over 2}}[\sin (\theta + 3\theta ) + \sin (\theta - 3\theta )]}}{{\sin \theta \cos \theta }}d\theta } \\ = \int {\,\frac{{\sin \theta \cos \theta + {\textstyle{1 \over 2}}(\sin 4\theta + \sin 2\theta ) - {\textstyle{1 \over 2}}(\sin 4\theta - \sin 2\theta {\rm{)}}}}{{\sin \theta \cos \theta }}d\theta } \\ = \int {\,\frac{{\sin \theta \cos \theta + \sin 2\theta }}{{\sin \theta \cos \theta }}d\theta } \\ = \int {\,\frac{{\sin \theta \cos \theta + 2\sin \theta \cos \theta }}{{\sin \theta \cos \theta }}d\theta } \\ = \int {\,3d\theta } \\ = \underline{\underline {3\theta + C}} \end{array}\)

A particle starts moving along the x-axis from\(x = 5\). Its velocity at any time t (in seconds) is given by \(v = 8t - 3\) where \(t \ge 0\).
(a) Find the velocity and acceleration, in magnitude and direction, when the particle starts to move.
(b) Express x in terms of t.
(c) Prove that the particle will not pass through the origin at any time.

(a) Let \(a{\rm{ u n i t s }}/{\rm{ }}{{\rm{s}}^2}\) be the acceleration of the particle at time t s.
\(\begin{array}{c}a = \frac{{dv}}{{dt}}\\ = 8\end{array}\)
When \(t = 0\),
\(\begin{array}{1}v = 8(0) - 3\\ = - 3\\a = 8\end{array}\)
∴ The required velocity is 3 units / s in the negative direction.
required acceleration is 8 units / s2 in the positive direction.

(b) ∵ \(\begin{array}{1}v = 8t - 3\\\frac{{dx}}{{dt}} = 8t - 3\end{array}\)
∴ \(\begin{array}{1}x = \int {\,(8t - 3)dt} \\ = 4{t^2} - 3t + C\end{array}\)
When \(t = 0\),\(x = 5\).
\(\begin{array}{1}5 = 4{(0)^2} - 3(0) + C\\C = 5\end{array}\)
∴ \(\underline{\underline {x = 4{t^2} - 3t + 5}} \)

(c) When the particle passes through the origin,\(x = 0\).
Consider\(4{t^2} - 3t + 5 = 0\),
\(\begin{array}{1}\Delta = {( - 3)^2} - 4(4)(5)\\ = - 71\\ < 0\end{array}\)
∴ The equation \(4{t^2} - 3t + 5 = 0\) has no real solutions.
∴ The particle will not pass through the origin.