Find the exact value of the integral ∫ 1 6 π 1 3 π sec 2 x tan 3 x d x , u = tan x

Let I = ∫ 1 6 π 1 3 π sec 2 x tan 3 x d x Put u = tan x ⇒ d u = sec 2 x d x When x = π 3 then u = 3 and when x = π 6 then u = 1 3 ∴ I = ∫ 1 3 3 u 3 d u = u 4 4 1 3 3 = 1 4 3 4 - 1 3 4 = 1 4 9 - 1 9 = 1 4 × 80 9 = 20 9

Find the exact value of the integral ∫ - 2 2 5 x 2 + 4 2 d x

Let I = ∫ - 2 2 5 x 2 + 4 2 d x Put x = 2 tan θ ⇒ d x = 2 sec 2 θ d θ When x = 2 then θ = π 4 and when x = - 2 then θ = - π 4 ∴ I = 10 ∫ - π 4 π 4 sec 2 θ 4 tan 2 θ + 4 2 d θ = 10 16 ∫ - π 4 π 4 sec 2 θ tan 2 θ + 1 2 d θ = 5 8 ∫ - π 4 π 4 sec 2 θ sec 4 θ d θ = 5 8 ∫ - π 4 π 4 1 sec 2 θ d θ = 5 8 ∫ - π 4 π 4 cos 2 θ d θ = 5 8 ∫ - π 4 π 4 cos 2 θ + 1 2 d θ ∵ cos 2 A = 2 cos 2 A - 1 = 5 16 sin 2 θ 2 + θ - π 4 π 4 = 5 16 1 2 sin π 2 - 1 2 sin - π 2 + π 4 + π 4 = 5 16 1 2 - - 1 2 + π 2 = 5 32 2 + π

Find the exact value of the integral ∫ 0 1 x 2 4 - x 2 d x

Let I = ∫ 0 1 x 2 4 - x 2 d x Put x = 2 cos θ ⇒ d x = - 2 sin θ d θ When x = 1 then θ = π 3 and when x = 0 then θ = π 2 ∴ I = - 2 ∫ π 2 π 3 4 sin θ cos 2 θ 4 - 4 cos 2 θ d θ = - 4 ∫ π 2 π 3 sin θ cos 2 θ 1 - cos 2 θ d θ = - 4 ∫ π 2 π 3 sin θ cos 2 θ sin θ d θ = - 4 ∫ π 2 π 3 cos 2 θ d θ = - 4 ∫ π 2 π 3 cos 2 θ + 1 2 d θ ∵ cos 2 A = 2 cos 2 A - 1 = - 2 sin 2 θ 2 + θ π 2 π 3 = - 2 1 2 sin 2 π 3 - 1 2 sin π + π 3 - π 2 = - 2 3 4 - 0 + π 3 - π 2 = - 2 3 3 + 4 π - 6 π 12 = 1 6 2 π - 3 3

Find the exact value of each of these integrals. ∫ 0 2 x 2 - x d x

Let I = ∫ 0 2 x ( 2 - x ) d x Put x = 2 sin 2 θ ⇒ d x = 4 sin θ cos θ d θ When x = 2 then θ = π 2 and when x = 0 then θ = 0 ∴ I = ∫ 0 π 2 2 sin 2 θ 2 - 2 sin 2 θ × 4 sin θ cos θ d θ = 8 ∫ 0 π 2 sin 2 θ 1 - sin 2 θ × sin θ cos θ d θ = 8 ∫ 0 π 2 sin 2 θ cos 2 θ × sin θ cos θ d θ = 8 ∫ 0 π 2 sin θ cos θ 2 d θ = 2 ∫ 0 π 2 sin 2 2 θ d θ ∵ sin 2 A = 2 sin A cos A = ∫ 0 π 2 1 - cos 4 θ d θ ∵ cos 2 A = 1 - 2 sin 2 A = θ - sin 4 θ 4 0 π 2 = π 2 - 0 - 1 4 sin 2 π + 1 4 sin 0 = π 2

Find the exact value of ∫ 1 2 4 1 + x - 2 2 d x .

Let I = ∫ 1 2 4 1 + ( x - 2 ) 2 d x Put u = x - 2 ⇒ d u = d x When x = 1 then u = - 1 and when x = 2 then u = 0 ∴ I = 4 ∫ - 1 0 d u u 2 + 1 = 4 tan - 1 u - 1 0 ∵ ∫ 1 1 + x 2 d x = tan − 1 ⁡ x + C = 4 tan - 1 ( 0 ) - tan - 1 ( - 1 ) = 4 0 - - π 4 = π

E M B I B E

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8D

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Further Calculus, Exercise 5: EXERCISE 8D

Author:Sue Pemberton, Julianne Hughes & Julian Gilbey

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Further Calculus, Exercise 5: EXERCISE 8D

Attempt the free practice questions on Chapter 8: Further Calculus, Exercise 5: EXERCISE 8D with hints and solutions to strengthen your understanding. Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book solutions are prepared by Experienced Embibe Experts.

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Further Calculus, Exercise 5: EXERCISE 8D with Hints & Solutions

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8D>Q 4

Find the exact value of the integral

$\int_{\frac{1}{6} π}^{\frac{1}{3} π} \sec^{2} x \tan^{3} x d x, u = \tan x$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8D>Q 4

Find the exact value of the integral

$\int_{- 2}^{2} \frac{5}{{(x^{2} + 4)}^{2}} d x$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8D>Q 4

Find the exact value of the integral

$\int_{0}^{1} \frac{x^{2}}{\sqrt{4 - x^{2}}} d x$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8D>Q 4

Find the exact value of each of these integrals.

$\int_{0}^{2} \sqrt{x (2 - x)} d x$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8D>Q 5

Find the exact value of $\int_{1}^{2} \frac{4}{1 + (x - 2)^{2}} d x .$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8D>Q 6

The diagram shows part of the curve $y = 2 \sin^{2} x \cos^{3} x .$ Using the substitution $u = \sin x,$ find the exact area of the shaded region.

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MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8D>Q 7

The diagram shows part of the curve $y = \frac{\ln x}{\sqrt{x}} .$ Using the substitution $u = \ln x,$ find the exact value of the volume of the solid formed when the shaded region is rotated completely about the $x$ -axis.

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MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8D>Q 8

The diagram shows part of the curve $y = 3 e^{\cos x} \sin x .$ The curve has rotational symmetry of order $2$ about the origin. Using the substitution $u = \cos x,$ find the exact value of the total area bounded by the curve and the $x$ -axis between $x = - π$ and $x = π .$

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