Integration by parts with inverse trig

5. Given that \(y = \operatorname { artanh } ( \cos x )\)

1. show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = - \operatorname { cosec } x$$
2. Hence find the exact value of $$\int _ { 0 } ^ { \frac { \pi } { 6 } } \cos x \operatorname { artanh } ( \cos x ) d x$$ giving your answer in the form \(a \ln ( b + c \sqrt { 3 } ) + d \pi\), where \(a , b , c\) and \(d\) are rational numbers to be found.
   (5)
   

8. $$y = \arccos ( 2 \sqrt { x } )$$

1. Determine \(\frac { \mathrm { d } y } { \mathrm {~d} x }\)
2. Show that $$\int y \mathrm {~d} x = x \arccos ( 2 \sqrt { x } ) + \int \frac { \sqrt { x } } { \sqrt { 1 - 4 x } } \mathrm {~d} x$$
3. Use the substitution \(\sqrt { x } = \frac { 1 } { 2 } \cos \theta\) to show that $$\int _ { 0 } ^ { \frac { 1 } { 8 } } \frac { \sqrt { x } } { \sqrt { 1 - 4 x } } \mathrm {~d} x = \frac { 1 } { 4 } \int _ { a } ^ { b } \cos ^ { 2 } \theta \mathrm {~d} \theta$$ where \(a\) and \(b\) are limits to be determined.
4. Hence, determine the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 8 } } \arccos ( 2 \sqrt { x } ) d x$$

4

1. Given that $$y = x \sqrt { 1 - x ^ { 2 } } - \cos ^ { - 1 } x$$ find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in a simplified form.
2. Hence, or otherwise, find the exact value of \(\int _ { 0 } ^ { 1 } 2 \sqrt { 1 - x ^ { 2 } } \mathrm {~d} x\).

5

1. Sb that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( x - \tan ^ { - 1 } x \right) = \frac { x ^ { 2 } } { 1 + x ^ { 2 } }\).
2. Sth the \(\int _ { 0 } ^ { \sqrt { 3 } } x \tan ^ { - 1 } x \mathrm {~d} x = \frac { 2 } { 3 } \pi - \frac { 1 } { 2 } \sqrt { 3 }\).

6 You are given that \(y = \tan ^ { - 1 } \sqrt { 2 x }\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Show that \(\int _ { \frac { 1 } { 6 } } ^ { \frac { 1 } { 2 } } \frac { \sqrt { x } } { \left( x + 2 x ^ { 2 } \right) } \mathrm { d } x = k \pi\) where \(k\) is a number to be determined in exact form.

4

1. Given that \(y = \tan ^ { - 1 } \sqrt { ( 3 x ) }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), giving your answer in terms of \(x\).
2. Hence, or otherwise, show that \(\int _ { \frac { 1 } { 3 } } ^ { 1 } \frac { 1 } { ( 1 + 3 x ) \sqrt { x } } \mathrm {~d} x = \frac { \sqrt { 3 } \pi } { n }\), where \(n\) is an integer.
   [0pt] [4 marks]

1. (a)

$$y = \tan ^ { - 1 } x$$ Assuming the derivative of \(\tan x\), prove that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 1 + x ^ { 2 } }$$ $$\mathrm { f } ( x ) = x \tan ^ { - 1 } 4 x$$ (b) Show that $$\int \mathrm { f } ( x ) \mathrm { d } x = A x ^ { 2 } \tan ^ { - 1 } 4 x + B x + C \tan ^ { - 1 } 4 x + k$$ where \(k\) is an arbitrary constant and \(A , B\) and \(C\) are constants to be determined.
(c) Hence find, in exact form, the mean value of \(\mathrm { f } ( x )\) over the interval \(\left[ 0 , \frac { \sqrt { 3 } } { 4 } \right]\)

4. Given that \(y = \operatorname { arsinh } ( \sqrt { } x ) , x > 0\),

1. find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), giving your answer as a simplified fraction.
2. Hence, or otherwise, find $$\int _ { \frac { 1 } { 4 } } ^ { 4 } \frac { 1 } { \sqrt { [ x ( x + 1 ) ] } } \mathrm { d } x$$ giving your answer in the form \(\ln \left( \frac { a + b \sqrt { } 5 } { 2 } \right)\), where \(a\) and \(b\) are integers.

5. (a) Given that \(y = \arctan 3 x\), and assuming the derivative of \(\tan x\), prove that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 } { 1 + 9 x ^ { 2 } }$$ (b) Show that $$\int _ { 0 } ^ { \frac { \sqrt { 3 } } { 3 } } 6 x \arctan 3 x \mathrm {~d} x = \frac { 1 } { 9 } ( 4 \pi - 3 \sqrt { } 3 )$$ (6)
[0pt] [P5 June 2002 Qn 6] \section*{6.} \begin{figure}[h] \end{figure} The curve \(C\) shown in Fig. 1 has equation \(y ^ { 2 } = 4 x , 0 \leq x \leq 1\).
The part of the curve in the first quadrant is rotated through \(2 \pi\) radians about the \(x\)-axis.
(a) Show that the surface area of the solid generated is given by $$4 \pi \int _ { 0 } ^ { 1 } \sqrt { ( 1 + x ) } d x$$ (b) Find the exact value of this surface area.
(c) Show also that the length of the curve \(C\), between the points \(( 1 , - 2 )\) and \(( 1,2 )\), is given by $$2 \int _ { 0 } ^ { 1 } \sqrt { \left( \frac { x + 1 } { x } \right) } \mathrm { d } x$$ (d) Use the substitution \(x = \sinh ^ { 2 } \theta\) to show that the exact value of this length is $$2 [ \sqrt { } 2 + \ln ( 1 + \sqrt { } 2 ) ]$$ [P5 June 2002 Qn 8]

11

1. Find \(\frac { \mathrm { d } } { \mathrm { d } x } \left( x ^ { 2 } \tan ^ { - 1 } x \right)\) 11
2. Hence find \(\int 2 x \tan ^ { - 1 } x \mathrm {~d} x\)

4 You are given that \(y = \tan ^ { - 1 } \sqrt { 2 x }\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Show that \(\int _ { \frac { 1 } { 6 } } ^ { \frac { 1 } { 2 } } \frac { \sqrt { x } } { \left( x + 2 x ^ { 2 } \right) } \mathrm { d } x = k \pi\) where \(k\) is a number to be determined in exact form.