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

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.

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.

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.

9

1. Given that \(x \geqslant 1\), use the substitution \(x = \cosh \theta\) to show that $$\int \frac { 1 } { x ^ { 2 } \sqrt { x ^ { 2 } - 1 } } \mathrm {~d} x = \frac { \sqrt { x ^ { 2 } - 1 } } { x } + C$$ where \(C\) is an arbitrary constant.
2. By differentiating sec \(y = x\) implicitly, show that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \sec ^ { - 1 } x \right) = \frac { 1 } { x \sqrt { x ^ { 2 } - 1 } }\) for \(x \geqslant 1\).
3. Use integration by parts to determine \(\int \frac { \sec ^ { - 1 } x } { x ^ { 2 } } \mathrm {~d} x\) for \(x \geqslant 1\).

In this question you must show detailed reasoning. Show that $$\int_0^{\frac{\pi}{3}} \operatorname{arcsinh} 2x \, dx = \frac{2}{3} \ln 3 - \frac{1}{3}.$$ [8]