Trigonometric substitution to simplify integral

5 Let \(I = \int _ { 0 } ^ { 1 } \frac { 9 } { \left( 3 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x\).

1. Using the substitution \(x = ( \sqrt { } 3 ) \tan \theta\), show that \(I = \sqrt { } 3 \int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \cos ^ { 2 } \theta \mathrm {~d} \theta\).
2. Hence find the exact value of \(I\).

5 Let \(I = \int _ { 0 } ^ { 1 } \frac { x ^ { 2 } } { \sqrt { } \left( 4 - x ^ { 2 } \right) } \mathrm { d } x\).

1. Using the substitution \(x = 2 \sin \theta\), show that $$I = \int _ { 0 } ^ { \frac { 1 } { 6 } \pi } 4 \sin ^ { 2 } \theta \mathrm {~d} \theta$$
2. Hence find the exact value of \(I\).

6 Let \(I = \int _ { 0 } ^ { 3 } \frac { 27 } { \left( 9 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x\).

1. Using the substitution \(x = 3 \tan \theta\), show that \(I = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \cos ^ { 2 } \theta \mathrm {~d} \theta\).
2. Hence find the exact value of \(I\).

11 Use the substitution \(2 x = \tan \theta\) to find the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 12 } { \left( 1 + 4 x ^ { 2 } \right) ^ { 2 } } d x$$ Give your answer in the form \(a + b \pi\), where \(a\) and \(b\) are rational numbers.
\includegraphics[max width=\textwidth, alt={}, center]{b1c4d339-322f-496d-833e-8b2d002d7c48-16_2716_38_143_2009}
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\includegraphics[max width=\textwidth, alt={}, center]{b1c4d339-322f-496d-833e-8b2d002d7c48-18_2715_35_143_2012}

5. $$I = \int \frac { 1 } { ( x - 1 ) \sqrt { } \left( x ^ { 2 } - 1 \right) } \mathrm { d } x , \quad x > 1$$ （a）Use the substitution \(x = 1 + u ^ { - 1 }\) to show that $$I = - \left( \frac { x + 1 } { x - 1 } \right) ^ { \frac { 1 } { 2 } } + c$$ （b）Hence show that $$\int _ { \sec \alpha } ^ { \sec \beta } \frac { 1 } { ( x - 1 ) \sqrt { } \left( x ^ { 2 } - 1 \right) } \mathrm { d } x = \cot \left( \frac { \alpha } { 2 } \right) - \cot \left( \frac { \beta } { 2 } \right) , \quad 0 < \alpha < \beta < \frac { \pi } { 2 }$$

6

1. Show that the substitution \(t = \tan \theta\) transforms the integral $$\int \frac { \mathrm { d } \theta } { 9 \cos ^ { 2 } \theta + \sin ^ { 2 } \theta }$$ into $$\int \frac { \mathrm { d } t } { 9 + t ^ { 2 } }$$
2. Hence show that $$\int _ { 0 } ^ { \frac { \pi } { 3 } } \frac { d \theta } { 9 \cos ^ { 2 } \theta + \sin ^ { 2 } \theta } = \frac { \pi } { 18 }$$

16 The point \(P ( 4,1,0 )\) is equidistant from the plane \(2 x + y + 2 z = 0\) and the line \(\frac { x - 3 } { 2 } = \frac { y - 1 } { b } = \frac { z + 5 } { 3 }\), where \(b > 0\). Determine the value of \(b\).

15 Three planes have equations $$\begin{aligned} x + k y + 3 z & = 1 \\ 3 x + 4 y + 2 z & = 3 \\ x + 3 y - z & = - k \end{aligned}$$ where \(k\) is a constant.

1. Show that the planes meet at a point except for one value of \(k\), which should be determined.
2. Show that, when the planes do meet at a point, the \(y\)-coordinate of this point is independent of \(k\).

15

1. Given that $$y = \operatorname { cosec } \theta$$ 15
2. 1. Express \(y\) in terms of \(\sin \theta\). 15
3. (ii) Hence, prove that $$\frac { \mathrm { d } y } { \mathrm {~d} \theta } = - \operatorname { cosec } \theta \cot \theta$$ 15
4. (iii) Show that $$\frac { \sqrt { y ^ { 2 } - 1 } } { y } = \cos \theta \quad \text { for } 0 < \theta < \frac { \pi } { 2 }$$ 15
5. 1. Use the substitution $$x = 2 \operatorname { cosec } u$$ to show that $$\int \frac { 1 } { x ^ { 2 } \sqrt { x ^ { 2 } - 4 } } \mathrm {~d} x \quad \text { for } x > 2$$ can be written as $$k \int \sin u \mathrm {~d} u$$ where \(k\) is a constant to be found.
      15
6. (ii) Hence, show $$\int \frac { 1 } { x ^ { 2 } \sqrt { x ^ { 2 } - 4 } } \mathrm {~d} x = \frac { \sqrt { x ^ { 2 } - 4 } } { 4 x } + c \quad \text { for } x > 2$$ where \(c\) is a constant.
   \includegraphics[max width=\textwidth, alt={}, center]{22ff390e-1360-43bd-8c7f-3d2b58627e91-32_2492_1721_217_150}
   \includegraphics[max width=\textwidth, alt={}]{22ff390e-1360-43bd-8c7f-3d2b58627e91-36_2496_1721_214_148}