Show substitution transforms integral, then evaluate

9. a. Use the substitution \(t ^ { 2 } = 2 x - 5\) to show that $$\int \frac { 1 } { x + 3 \sqrt { 2 x - 5 } } \mathrm {~d} x = \int \frac { 2 t } { t ^ { 2 } + 6 t + 5 } \mathrm {~d} t$$ b. Hence find the exact value of $$\int _ { 3 } ^ { 27 } \frac { 1 } { x + 3 \sqrt { 2 x - 5 } } \mathrm {~d} x$$

1. (a) Use the substitution \(u = 1 + \sqrt { x }\) to show that

$$\int _ { 0 } ^ { 16 } \frac { \mathrm { x } } { 1 + \sqrt { \mathrm { x } } } \mathrm {~d} x = \int _ { p } ^ { q } \frac { 2 ( u - 1 ) ^ { 3 } } { u } \mathrm {~d} u$$ where \(p\) and \(q\) are constants to be found.
(b) Hence show that $$\int _ { 0 } ^ { 16 } \frac { \mathrm { x } } { 1 + \sqrt { \mathrm { x } } } \mathrm {~d} x = A - B \ln 5$$ where \(A\) and \(B\) are constants to be found.

8. $$f ( x ) = \frac { 3 } { 13 + 6 \sin x - 5 \cos x }$$ Using the substitution \(t = \tan \left( \frac { x } { 2 } \right)\)

1. show that \(\mathrm { f } ( x )\) can be written in the form $$\frac { 3 \left( 1 + t ^ { 2 } \right) } { 2 ( 3 t + 1 ) ^ { 2 } + 6 }$$
2. Hence solve, for \(0 < x < 2 \pi\), the equation $$\mathrm { f } ( x ) = \frac { 3 } { 7 }$$ giving your answers to 2 decimal places where appropriate.
3. Use the result of part (a) to show that $$\int _ { \frac { \pi } { 3 } } ^ { \frac { 4 \pi } { 3 } } f ( x ) d x = K \left( \arctan \left( \frac { \sqrt { 3 } - 9 } { 3 } \right) - \arctan \left( \frac { \sqrt { 3 } + 3 } { 3 } \right) + \pi \right)$$ where \(K\) is a constant to be determined.

1. (a) Show that the substitution \(t = \tan \left( \frac { x } { 2 } \right)\) transforms the integral

$$\int \frac { 1 } { 2 \sin x - \cos x + 5 } d x$$ into the integral $$\int \frac { 1 } { 3 t ^ { 2 } + 2 t + 2 } \mathrm {~d} t$$ (b) Hence determine $$\int \frac { 1 } { 2 \sin x - \cos x + 5 } d x$$