Integration using inverse trig and hyperbolic functions

Evaluate the integral $$\int_0^1 \frac{dx}{\sqrt{2x^2 + 4x + 6}}.$$ [6]

2. (a) Find $$\int \frac { 1 } { \sqrt { } \left( 4 x ^ { 2 } + 9 \right) } d x$$ (b) Use your answer to part (a) to find the exact value of $$\int _ { - 3 } ^ { 3 } \frac { 1 } { \sqrt { \left( 4 x ^ { 2 } + 9 \right) } } d x$$ giving your answer in the form \(k \ln ( a + b \sqrt { } 5 )\), where \(a\) and \(b\) are integers and \(k\) is a constant.

1. The curve \(C\) has equation

$$y = \frac { 1 } { \sqrt { x ^ { 2 } + 2 x - 3 } } , \quad x > 1$$

1. Find \(\int y \mathrm {~d} x\) The region \(R\) is bounded by the curve \(C\), the \(x\)-axis and the lines with equations \(x = 2\) and \(x = 3\). The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
2. Find the volume of the solid generated. Give your answer in the form \(p \pi \ln q\), where \(p\) and \(q\) are rational numbers to be found.

1 Find the exact value of \(\int _ { 2 } ^ { \frac { 7 } { 2 } } \frac { 1 } { \sqrt { 4 x - x ^ { 2 } - 1 } } \mathrm {~d} x\).

2. Determine

1. \(\int \frac { 1 } { 3 x ^ { 2 } + 12 x + 24 } \mathrm {~d} x\)
2. \(\int \frac { 1 } { \sqrt { 27 - 6 x - x ^ { 2 } } } \mathrm {~d} x\)

Using calculus, find the exact values of

1. \(\int_1^2 \frac{1}{x^2 - 4x + 5} \, dx\) [3]
2. \(\int_{\sqrt{3}}^3 \frac{\sqrt{x^2 - 3}}{x^2} \, dx\) [5]

1. Find \(\frac{d}{dx}(x^2\tan^{-1} x)\) [1 mark]
2. Hence find \(\int 2x \tan^{-1} x \, dx\) [4 marks]

1. Find \(\frac{d}{dx}(x^2\tan^{-1} x)\) [1 mark]
2. Hence find \(\int 2x \tan^{-1} x \, dx\) [4 marks]

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)
   

Use a suitable trigonometric substitution to find \(\int \frac{x^2}{\sqrt{1-x^2}} \text{d}x\). [7]

1. By differentiating \(\frac{1}{\cos x}\), show that the derivative of \(\sec x\) is \(\sec x \tan x\). Hence show that if \(y = \ln(\sec x + \tan x)\) then \(\frac{dy}{dx} = \sec x\). [4]
2. Using the substitution \(x = (\sqrt{3}) \tan \theta\), find the exact value of $$\int_1^3 \frac{1}{\sqrt{(3 + x^2)}} dx,$$ expressing your answer as a single logarithm. [4]

5. $$I = \int \frac { 1 } { ( x - 1 ) \sqrt { } \left( x ^ { 2 } - 1 \right) } \mathrm { d } x , \quad x > 1$$

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

Given that \(y = \arctan \frac{x}{\sqrt{1 + x^2}}\) show that \(\frac{dy}{dx} = \frac{1}{\sqrt{1 + x^2}}\) [4]

2

1. Given that \(y = \tan ^ { - 1 } x\), prove that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 1 + x ^ { 2 } }\).
2. Verify that \(y = \tan ^ { - 1 } x\) satisfies the equation $$\left( 1 + x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 0$$

13

1. (a) Given that \(x \geqslant 1\), show that \(\sec ^ { - 1 } x = \cos ^ { - 1 } \left( \frac { 1 } { x } \right)\), and deduce that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \sec ^ { - 1 } x \right) = \frac { 1 } { x \sqrt { x ^ { 2 } - 1 } }\).
   (b) Use integration by parts to determine \(\int \sec ^ { - 1 } x \mathrm {~d} x\).
2. \includegraphics[max width=\textwidth, alt={}, center]{5d526fd9-72f8-42b1-b156-fd4a0c764c82-4_670_1029_1073_596} The diagram shows the curve \(S\) with equation \(y = \sec ^ { - 1 } x\) for \(x \geqslant 1\). The line \(L\), with gradient \(\frac { 1 } { \sqrt { 2 } }\), is the tangent to \(S\) at the point \(P\) and cuts the \(x\)-axis at the point \(Q\). The point \(I\) has coordinates \(( 1,0 )\).
   (a) Determine the exact coordinates of \(P\) and \(Q\).
   (b) The region \(R\), shaded on the diagram, is bounded by the line segments \(P Q\) and \(Q I\) and the \(\operatorname { arc } I P\) of \(S\). Show that \(R\) has area $$\ln ( 1 + \sqrt { 2 } ) - \frac { \pi ( 8 - \pi ) \sqrt { 2 } } { 32 } .$$ {www.cie.org.uk} after the live examination series. }

2 Find the exact value of \(\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 3 + 4 x - 4 x ^ { 2 } } } \mathrm {~d} x\).

In this question you must show detailed reasoning. Find \(\int_{2}^{\infty} \frac{1}{4+x^2} \, dx\). [4]

Find \(\int \sqrt{x^2 + 4} \, dx\). (Total 7 marks)

2. Use calculus to find the exact value of \(\int _ { - 2 } ^ { 1 } \frac { 1 } { x ^ { 2 } + 4 x + 13 } \mathrm {~d} x\).

Show that \(\int_0^{\frac{1}{\sqrt{3}}} \frac{4}{1-x^4} dx = \ln(a + \sqrt{b}) + \frac{\pi}{c}\) where \(a\), \(b\) and \(c\) are integers to be determined. [6]

1. Prove that the derivative of \(\operatorname{artanh} x\), \(-1 < x < 1\), is \(\frac{1}{1-x^2}\). [3]
2. Find \(\int \operatorname{artanh} x \, dx\). [4]

Find the value of \(\int_6^7 \frac{1}{\sqrt{(x-5)^2-1}} \, dx\), giving your answer in the form \(\ln(a + \sqrt{b})\), where \(a\) and \(b\) are integers to be determined. [4]

3 Use the substitution \(t = \tan \frac { 1 } { 2 } x\) to show that $$\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \frac { 1 } { 1 - \sin x } \mathrm {~d} x = 1 + \sqrt { 3 }$$

6 [In this question you may use, without proof, the formula \(\int \sec x \mathrm {~d} x = \ln ( \sec x + \tan x ) + \operatorname { const }\).]

1. Let \(y = \sec x\). Find the mean value of \(y\) with respect to \(x\) over the interval \(\frac { 1 } { 6 } \pi \leqslant x \leqslant \frac { 1 } { 3 } \pi\).
2. The curve \(C\) has equation \(y = - \ln ( \cos x )\), for \(0 \leqslant x \leqslant \frac { 1 } { 3 } \pi\). Find the arc length of \(C\).

4. The curve \(C\) has equation $$y = \operatorname { arsinh } x + x \sqrt { x ^ { 2 } + 1 } , \quad 0 \leqslant x \leqslant 1$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \sqrt { x ^ { 2 } + 1 }\)
2. Hence show that the length of the curve \(C\) is given by $$\int _ { 0 } ^ { 1 } \sqrt { 4 x ^ { 2 } + 5 } d x$$
3. Using the substitution \(x = \frac { \sqrt { 5 } } { 2 } \sinh u\), find the exact length of the curve \(C\), giving your answer in the form \(a + b \ln c\), where \(a , b\) and \(c\) are constants to be found.

7

1. Given that \(y = \tan ^ { - 1 } \left( \frac { 1 + x } { 1 - x } \right)\) and \(x \neq 1\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 1 + x ^ { 2 } }\).
   [0pt] [4 marks]
2. Hence, given that \(x < 1\), show that \(\tan ^ { - 1 } \left( \frac { 1 + x } { 1 - x } \right) - \tan ^ { - 1 } x = \frac { \pi } { 4 }\).
   [0pt] [3 marks]