Chapter 4 - Integration - 4.9 Evaluating Definite Integrals By Substitution - Exercises Set 4.9 - Page 340: 1

$$\left( {\bf{a}} \right)\frac{1}{2}\int_1^5 {{u^3}du,} \left( {\bf{b}} \right) - \frac{3}{2}\int_{25}^9 {\sqrt u du,} \left( {\bf{c}} \right)\frac{1}{\pi }\int_{ - \pi /2}^{\pi /2} {\cos udu} ,\left( {\bf{d}} \right)\int_1^2 {\left( {u + 1} \right){u^5}} du$$

$$\eqalign{ & \left( {\bf{a}} \right)\int_1^3 {{{\left( {2x - 1} \right)}^3}dx} ,{\text{ }} \cr & {\text{Set }}u = 2x - 1,\,\,du = 2dx \to \,\,\,dx = \frac{1}{2}du \cr & {\text{if }}x = 1,\,\,\,u = 2\left( 1 \right) - 1,\,\,u = 1 \cr & {\text{if }}x = 3,\,\,\,u = 2\left( 3 \right) - 1,\,\,u = 5 \cr & {\text{Thus}}{\text{,}} \cr & \,\,\,\,\,\int_1^3 {{{\left( {2x - 1} \right)}^3}dx} = \int_1^5 {{u^3}\left( {\frac{1}{2}} \right)du} \cr & \,\,\,\,\, = \frac{1}{2}\int_1^5 {{u^3}du} \cr & \cr & \left( {\bf{b}} \right)\int_0^4 {3x\sqrt {25 - {x^2}} dx} ,{\text{ }} \cr & {\text{Set }}u = 25 - {x^2},\,\,du = - 2xdx \to \,\,\,xdx = - \frac{1}{2}du \cr & {\text{if }}x = 0,\,\,\,u = 25 - {\left( 0 \right)^2},\,\,u = 25 \cr & {\text{if }}x = 4,\,\,\,u = 25 - {\left( 4 \right)^2},\,\,u = 9 \cr & {\text{Thus}}{\text{,}} \cr & \,\,\int_0^4 {3x\sqrt {25 - {x^2}} dx} = - \frac{3}{2}\int_{25}^9 {3\sqrt u \left( { - \frac{1}{2}} \right)du} \cr & \,\,\,\,\, = - \frac{3}{2}\int_{25}^9 {\sqrt u du} \cr & \cr & \left( {\bf{c}} \right)\int_{ - 1/2}^{1/2} {\cos \left( {\pi \theta } \right)d\theta } ;\,\,u = \pi \theta {\text{ }} \cr & {\text{Set }}u = \pi \theta ,\,\,du = \pi d\theta \to \,\,\,d\theta = \frac{1}{\pi }du \cr & {\text{if }}\theta = 1/2,\,\,\,u = \pi \left( {1/2} \right),\,\,u = \pi /2 \cr & {\text{if }}\theta = - 1/2,\,\,\,u = \pi \left( { - 1/2} \right),\,\,u = - \pi /2 \cr & {\text{Thus}}{\text{,}} \cr & \,\,\int_{ - 1/2}^{1/2} {\cos \left( {\pi \theta } \right)d\theta } = - \frac{3}{2}\int_{ - \pi /2}^{\pi /2} {\cos u\left( {\frac{1}{\pi }} \right)du} \cr & \,\,\,\,\, = \frac{1}{\pi }\int_{ - \pi /2}^{\pi /2} {\cos udu} \cr & \cr & \left( {\bf{d}} \right)\int_0^1 {\left( {x + 2} \right){{\left( {x + 1} \right)}^5}dx} ;\,\,u = x + 1 \cr & {\text{Set }}u = x + 1,\,\,du = dx \cr & {\text{if }}x = 1,\,\,\,u = 1 + 1,\,\,u = 2 \cr & {\text{if }}x = 0,\,\,\,u = 0 + 1,\,\,u = 1 \cr & {\text{Thus}}{\text{,}} \cr & \,\,\int_0^1 {\left( {x + 2} \right){{\left( {x + 1} \right)}^5}dx} = \int_1^2 {\left( {u + 1} \right){u^5}} du \cr} $$