Chapter 4 - Integration - 4.9 Evaluating Definite Integrals By Substitution - Exercises Set 4.9 - Page 340: 2

$$\eqalign{ & \left( {\bf{a}} \right) - \frac{1}{2}\int_7^{ - 3} {{u^8}du} \cr & \left( {\bf{b}} \right)\int_{5/2}^{3/2} {\frac{{du}}{{\sqrt u }}} \cr & \left( {\bf{c}} \right)\int_0^1 u du \cr & \left( {\bf{d}} \right)\frac{1}{2}\int_3^4 {\left( {{u^{3/2}} - 3{u^{1/2}}} \right)du} \cr} $$

$$\eqalign{ & \left( {\bf{a}} \right)\int_{ - 1}^4 {{{\left( {5 - 2x} \right)}^8}dx;{\text{ }}u = 5 - 2x} \cr & {\text{Apply the substitution}} \cr & u = 5 - 2x \Rightarrow {\left( {5 - 2x} \right)^8} = {u^8} \cr & du = - 2dx \Rightarrow dx = - \frac{1}{2}du \cr & {\text{The new limits of integration are:}} \cr & x = - 1 \Rightarrow u = 5 - 2\left( { - 1} \right) = 7 \cr & x = 4{\text{ }} \Rightarrow u = 5 - 2\left( 4 \right) = - 3 \cr & \int_{ - 1}^4 {{{\left( {5 - 2x} \right)}^8}dx} = - \frac{1}{2}\int_7^{ - 3} {{u^8}du} \cr & \cr & \left( {\bf{b}} \right)\int_{ - \pi /3}^{2\pi /3} {\frac{{\sin x}}{{\sqrt {2 + \cos x} }}dx;{\text{ }}u = 2 + \cos x} ,{\text{ }}du = \sin xdx \cr & {\text{The new limits of integration are:}} \cr & x = 2\pi /3{\text{ }} \Rightarrow u = 2 + \cos \left( {\frac{{2\pi }}{3}} \right) = \frac{3}{2} \cr & x = - \pi /3{\text{ }} \Rightarrow u = 2 + \cos \left( { - \frac{\pi }{3}} \right) = \frac{5}{2} \cr & {\text{Apply the substitution}} \cr & \int_{ - \pi /3}^{2\pi /3} {\frac{{\sin x}}{{\sqrt {2 + \cos x} }}dx} = \int_{5/2}^{3/2} {\frac{{du}}{{\sqrt u }}} \cr & \cr & \left( {\bf{c}} \right)\int_0^{\pi /4} {{{\tan }^2}x{{\sec }^2}xdx;{\text{ }}u = \tan x} ,{\text{ }}du = {\sec ^2}xdx \cr & {\text{The new limits of integration are:}} \cr & x = \pi /4{\text{ }} \Rightarrow u = \tan \left( {\frac{\pi }{4}} \right) = 1 \cr & x = 0{\text{ }} \Rightarrow u = \tan \left( 0 \right) = 0 \cr & {\text{Apply the substitution}} \cr & \int_0^{\pi /4} {{{\tan }^2}x{{\sec }^2}xdx} = \int_0^1 u du \cr & \cr & \left( {\bf{d}} \right)\int_0^1 {{x^3}\sqrt {{x^2} + 3} dx} ,{\text{ }}u = {x^2} + 3,{\text{ }}du = 2xdx,{\text{ }}xdx = \frac{1}{2}du \cr & {\text{The new limits of integration are:}} \cr & x = 1{\text{ }} \Rightarrow u = {\left( 1 \right)^2} + 3 = 4 \cr & x = 0{\text{ }} \Rightarrow u = {\left( 0 \right)^2} + 3 = 3 \cr & {\text{Apply the substitution}} \cr & \int_0^1 {{x^3}\sqrt {{x^2} + 3} dx} = \int_0^1 {{x^2}\sqrt {{x^2} + 3} \left( x \right)dx} \cr & {\text{ }} = \int_3^4 {\left( {u - 3} \right)\sqrt u \left( {\frac{1}{2}} \right)du} \cr & {\text{ }} = \frac{1}{2}\int_3^4 {\left( {{u^{3/2}} - 3{u^{1/2}}} \right)du} \cr} $$