Showing integral equals given value

7 Let \(I = \int _ { 1 } ^ { 4 } \frac { 1 } { x ( 4 - \sqrt { } x ) } \mathrm { d } x\).

1. Use the substitution \(u = \sqrt { } x\) to show that \(I = \int _ { 1 } ^ { 2 } \frac { 2 } { u ( 4 - u ) } \mathrm { d } u\).
2. Hence show that \(I = \frac { 1 } { 2 } \ln 3\).

8 Let \(I = \int _ { 2 } ^ { 5 } \frac { 5 } { x + \sqrt { } ( 6 - x ) } \mathrm { d } x\).

1. Using the substitution \(u = \sqrt { } ( 6 - x )\), show that $$I = \int _ { 1 } ^ { 2 } \frac { 10 u } { ( 3 - u ) ( 2 + u ) } \mathrm { d } u$$
2. Hence show that \(I = 2 \ln \left( \frac { 9 } { 2 } \right)\).

6 Let \(I = \int _ { 0 } ^ { 1 } \frac { \sqrt { } x } { 2 - \sqrt { } x } \mathrm {~d} x\).

1. Using the substitution \(u = 2 - \sqrt { } x\), show that \(I = \int _ { 1 } ^ { 2 } \frac { 2 ( 2 - u ) ^ { 2 } } { u } \mathrm {~d} u\).
2. Hence show that \(I = 8 \ln 2 - 5\).

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

1. Using the substitution \(u = 1 + x ^ { 2 }\), show that \(I = \int _ { 1 } ^ { 2 } \frac { ( u - 1 ) ^ { 2 } } { 2 u ^ { 3 } } \mathrm {~d} u\).
2. Hence find the exact value of \(I\).

10 By first using the substitution \(u = \mathrm { e } ^ { x }\), show that $$\int _ { 0 } ^ { \ln 4 } \frac { \mathrm { e } ^ { 2 x } } { \mathrm { e } ^ { 2 x } + 3 \mathrm { e } ^ { x } + 2 } \mathrm {~d} x = \ln \left( \frac { 8 } { 5 } \right)$$

5

1. Prove the identity \(\tan 2 \theta - \tan \theta \equiv \tan \theta \sec 2 \theta\).
2. Hence show that \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \tan \theta \sec 2 \theta \mathrm {~d} \theta = \frac { 1 } { 2 } \ln \frac { 3 } { 2 }\).

6 Let \(I = \int _ { 1 } ^ { 4 } \frac { ( \sqrt { } x ) - 1 } { 2 ( x + \sqrt { } x ) } \mathrm { d } x\).

1. Using the substitution \(u = \sqrt { } x\), show that \(I = \int _ { 1 } ^ { 2 } \frac { u - 1 } { u + 1 } \mathrm {~d} u\).
2. Hence show that \(I = 1 + \ln \frac { 4 } { 9 }\).

7

1. Use the substitution \(u = \cos x\) to show that $$\int _ { 0 } ^ { \pi } \sin 2 x \mathrm { e } ^ { 2 \cos x } \mathrm {~d} x = \int _ { - 1 } ^ { 1 } 2 u \mathrm { e } ^ { 2 u } \mathrm {~d} u$$
2. Hence find the exact value of \(\int _ { 0 } ^ { \pi } \sin 2 x \mathrm { e } ^ { 2 \cos x } \mathrm {~d} x\).

3. (i) Find, in simplest form, $$\int ( 2 x - 5 ) ^ { 7 } \mathrm {~d} x$$ (ii) Show, by algebraic integration, that $$\int _ { 0 } ^ { \frac { \pi } { 3 } } \frac { 4 \sin x } { 1 + 2 \cos x } \mathrm {~d} x = \ln a$$ where \(a\) is a rational constant to be found.

1. Using the substitution \(u = \cos x + 1\), or otherwise, show that

$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \mathrm { e } ^ { ( \cos x + 1 ) } \sin x \mathrm {~d} x = \mathrm { e } ( \mathrm { e } - 1 )$$

5. In this question you should show all stages of your working. Solutions relying on calculator technology are not acceptable.
Using the substitution \(u = 3 + \sqrt { 2 x - 1 }\) find the exact value of $$\int _ { 1 } ^ { 13 } \frac { 4 } { 3 + \sqrt { 2 x - 1 } } d x$$ giving your answer in the form \(p + q \ln 2\), where \(p\) and \(q\) are integers to be found.

2. Using the substitution \(u = \cos x + 1\), or otherwise, show that $$\int _ { 0 } ^ { \frac { \pi } { 2 } } \mathrm { e } ^ { \cos x + 1 } \sin x \mathrm {~d} x = \mathrm { e } ( \mathrm { e } - 1 )$$ (6)

3. Using the substitution \(u = 2 + \sqrt { } ( 2 x + 1 )\), or other suitable substitutions, find the exact value of $$\int _ { 0 } ^ { 4 } \frac { 1 } { 2 + \sqrt { } ( 2 x + 1 ) } d x$$ giving your answer in the form \(A + 2 \ln B\), where \(A\) is an integer and \(B\) is a positive constant.

5 Using the substitution \(u = 2 x + 1\), show that \(\int _ { 0 } ^ { 1 } \frac { x } { 2 x + 1 } \mathrm {~d} x = \frac { 1 } { 4 } ( 2 - \ln 3 )\).

4 Show that \(\int _ { 1 } ^ { 4 } \frac { x } { x ^ { 2 } + 2 } \mathrm {~d} x = \frac { 1 } { 2 } \ln 6\).

5 Using a suitable substitution or otherwise, show that \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { \sin 2 x } { 3 + \cos 2 x } \mathrm {~d} x = \frac { 1 } { 2 } \ln 2\).

1 Show that \(\int _ { 1 } ^ { 2 } \frac { 1 } { \sqrt { 3 x - 2 } } \mathrm {~d} x = \frac { 2 } { 3 }\).

5 Show that \(\int _ { 1 } ^ { 4 } \frac { x } { x ^ { 2 } + 2 } \mathrm {~d} x = \frac { 1 } { 2 } \ln 6\).

4 Use the substitution \(u = 2 x - 5\) to show that \(\int _ { \frac { 5 } { 2 } } ^ { 3 } ( 4 x - 8 ) ( 2 x - 5 ) ^ { 7 } \mathrm {~d} x = \frac { 17 } { 72 }\).

3. Using the substitution \(u = \mathrm { e } ^ { x } - 1\), show that $$\int _ { \ln 2 } ^ { \ln 5 } \frac { \mathrm { e } ^ { 2 x } } { \sqrt { \mathrm { e } ^ { x } - 1 } } \mathrm {~d} x = \frac { 20 } { 3 }$$

6 Using a suitable substitution or otherwise, show that \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { \sin 2 x } { 3 + \cos 2 x } \mathrm {~d} x = \frac { 1 } { 2 } \ln 2\).

6 Use the substitution \(u = 1 + \sqrt { x }\) to show that $$\int _ { 4 } ^ { 9 } \frac { 1 } { 1 + \sqrt { x } } \mathrm {~d} x = 2 + 2 \ln \frac { 3 } { 4 }$$

9 Use the substitution \(x = 2 \sin \theta\) to show that \(\int _ { 1 } ^ { \sqrt { 3 } } \sqrt { 4 - x ^ { 2 } } \mathrm {~d} x = \frac { 1 } { 3 } \pi\).

5

1. Use the trapezium rule, with two strips of equal width, to show that $$\int _ { 0 } ^ { 4 } \frac { 1 } { 2 + \sqrt { x } } \mathrm {~d} x \approx \frac { 11 } { 4 } - \sqrt { 2 }$$
2. Use the substitution \(x = u ^ { 2 }\) to find the exact value of $$\int _ { 0 } ^ { 4 } \frac { 1 } { 2 + \sqrt { x } } \mathrm {~d} x$$
3. Using your answers to parts (i) and (ii), show that $$\ln 2 \approx k + \frac { \sqrt { 2 } } { 4 }$$ where \(k\) is a rational number to be determined.

1. Show that

$$\int _ { 0 } ^ { 2 } 2 x \sqrt { x + 2 } \mathrm {~d} x = \frac { 32 } { 15 } ( 2 + \sqrt { 2 } )$$