Show integral transforms via substitution then evaluate (trigonometric/Weierstrass)

6

1. Use the substitution \(x = \sin ^ { 2 } \theta\) to show that $$\int \sqrt { } \left( \frac { x } { 1 - x } \right) \mathrm { d } x = \int 2 \sin ^ { 2 } \theta \mathrm {~d} \theta$$
2. Hence find the exact value of $$\left. \int _ { 0 } ^ { \frac { 1 } { 4 } } \sqrt { ( } \frac { x } { 1 - x } \right) \mathrm { d } x$$

6. (i) Given that \(y > 0\), find $$\int \frac { 3 y - 4 } { y ( 3 y + 2 ) } d y$$ (ii) (a) Use the substitution \(x = 4 \sin ^ { 2 } \theta\) to show that $$\int _ { 0 } ^ { 3 } \sqrt { \left( \frac { x } { 4 - x } \right) } \mathrm { d } x = \lambda \int _ { 0 } ^ { \frac { \pi } { 3 } } \sin ^ { 2 } \theta \mathrm {~d} \theta$$ where \(\lambda\) is a constant to be determined.
(b) Hence use integration to find $$\int _ { 0 } ^ { 3 } \sqrt { \left( \frac { x } { 4 - x } \right) } d x$$ giving your answer in the form \(a \pi + b\), where \(a\) and \(b\) are exact constants.

6

1. Show that the substitution \(x = \sin ^ { 2 } \theta\) transforms \(\int \sqrt { \frac { x } { 1 - x } } \mathrm {~d} x\) to \(\int 2 \sin ^ { 2 } \theta \mathrm {~d} \theta\).
2. Hence find \(\int _ { 0 } ^ { 1 } \sqrt { \frac { x } { 1 - x } } \mathrm {~d} x\).

5 It is given that \(I = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { \cos \theta } { 1 + \cos \theta } \mathrm { d } \theta\).

1. By using the substitution \(t = \tan \frac { 1 } { 2 } \theta\), show that \(I = \int _ { 0 } ^ { 1 } \left( \frac { 2 } { 1 + t ^ { 2 } } - 1 \right) \mathrm { d } t\).
2. Hence find \(I\) in terms of \(\pi\).

1. (a) Given that \(a\) is a positive constant, use the substitution \(x = a \sin ^ { 2 } \theta\) to show that

$$\int _ { 0 } ^ { a } x ^ { \frac { 1 } { 2 } } \sqrt { a - x } \mathrm {~d} x = \frac { 1 } { 2 } a ^ { 2 } \int _ { 0 } ^ { \frac { \pi } { 2 } } \sin ^ { 2 } 2 \theta \mathrm {~d} \theta$$ (b) Hence use algebraic integration to show that $$\int _ { 0 } ^ { a } x ^ { \frac { 1 } { 2 } } \sqrt { a - x } d x = k \pi a ^ { 2 }$$ where \(k\) is a constant to be found.

1. In this question you must show all stages of your working.

\section*{Solutions relying on calculator technology are not acceptable.}

1. Use the substitution \(t = \tan \left( \frac { \theta } { 2 } \right)\) to show that $$\int \frac { 1 } { 2 \sin \theta + \cos \theta + 2 } d \theta = \int \frac { a } { ( t + b ) ^ { 2 } + c } d t$$ where \(a\), \(b\) and \(c\) are constants to be determined.
2. Hence show that $$\int _ { \frac { \pi } { 2 } } ^ { \frac { 2 \pi } { 3 } } \frac { 1 } { 2 \sin \theta + \cos \theta + 2 } d \theta = \ln \left( \frac { 2 \sqrt { 3 } } { 3 } \right)$$

1. Show that the two non-stationary points of inflection on the curve \(y = \ln(1 + 4x^2)\) are at \(x = \pm\frac{1}{2}\). [6]

\includegraphics{figure_9} The diagram shows the curve \(y = \ln(1 + 4x^2)\). The shaded region is bounded by the curve and a line parallel to the \(x\)-axis which meets the curve where \(x = \frac{1}{2}\) and \(x = -\frac{1}{2}\).

1. Show that the area of the shaded region is given by $$\int_0^{\ln 2} \sqrt{e^y - 1} \, dy.$$ [3]
2. Show that the substitution \(e^y = \sec^2\theta\) transforms the integral in part (ii) to \(\int_0^{\frac{\pi}{4}} 2\tan^2\theta \, d\theta\). [2]
3. Hence find the exact area of the shaded region. [3]