Substitution then integration by parts

7 The integral \(I\) is defined by \(I = \int _ { 0 } ^ { 2 } 4 t ^ { 3 } \ln \left( t ^ { 2 } + 1 \right) \mathrm { d } t\).

1. Use the substitution \(x = t ^ { 2 } + 1\) to show that \(I = \int _ { 1 } ^ { 5 } ( 2 x - 2 ) \ln x \mathrm {~d} x\).
2. Hence find the exact value of \(I\).

1. (a) Using the substitution \(u = \sqrt { 2 x + 1 }\), show that

$$\int _ { 4 } ^ { 12 } \sqrt { 8 x + 4 } \mathrm { e } ^ { \sqrt { 2 x + 1 } } \mathrm {~d} x$$ may be expressed in the form $$\int _ { a } ^ { b } k u ^ { 2 } \mathrm { e } ^ { u } \mathrm {~d} u$$ where \(a\), \(b\) and \(k\) are constants to be found.
(b) Hence find, by algebraic integration, the exact value of $$\int _ { 4 } ^ { 12 } \sqrt { 8 x + 4 } e ^ { \sqrt { 2 x + 1 } } d x$$ giving your answer in simplest form.

7

1. Use integration by parts to find \(\int ( t - 1 ) \ln t \mathrm {~d} t\).
2. Use the substitution \(t = 2 x + 1\) to show that \(\int 4 x \ln ( 2 x + 1 ) \mathrm { d } x\) can be written as \(\int ( t - 1 ) \ln t \mathrm {~d} t\).
3. Hence find the exact value of \(\int _ { 0 } ^ { 1 } 4 x \ln ( 2 x + 1 ) \mathrm { d } x\).

8. A curve, which is part of an ellipse, has parametric equations $$x = 3 \cos \theta , \quad y = 5 \sin \theta , \quad 0 \leq \theta \leq \frac { \pi } { 2 }$$ The curve is rotated through \(2 \pi\) radians about the \(x\)-axis.

1. Show that the area of the surface generated is given by the integral $$k \pi \int _ { 0 } ^ { a } \sqrt { } \left( 16 c ^ { 2 } + 9 \right) \mathrm { d } c , \text { where } c = \cos \theta$$ and where \(k\) and \(\alpha\) are constants to be found.
2. Using the substitution \(c = \frac { 3 } { 4 } \sinh u\), or otherwise, evaluate the integral, showing all of your working and giving the final answer to 3 significant figures.