Chapter 5 More on integration

Exercise 10.1

Q1

(a)\[ \int 3\sec^2{x} \ dx = 3 \int \sec^2{x} \ dx = 3 \tan{x}+C \]

(b)\[ \int (2 e^x+\cos{x}) \ dx = \int 2e^x \ dx + \int \cos{x} \ dx \\ = 2e^x+\sin{x}+C \]

(c)\[ \int \frac{2x+3}{x} \ dx = \int (2+\frac{3}{x}) \ dx \\ = \int 2 \ dx + \int \frac{3}{x} \ dx \\ = 2x+3\ln{\mid x \mid}+C \]

(d)\[ \int (1+\frac{1}{x})^2 \ dx = \int (1+\frac{2}{x}+\frac{1}{x^2}) \ dx \\ = \int 1 \ dx + \int \frac{2}{x} \ dx + \int \frac{1}{x^2} \ dx \\ = x+2\ln{\mid x \mid}-\frac{1}{x}+C \]

Q2

(a)\[ \int_0^{\frac{\pi}{3}} 2\sin{x} \ dx = (-2\cos{x})\mid_0^{\frac{\pi}{3}} \\ = -2\cos{\frac{\pi}{3}}-(-2\cos{0}) \\ = -2 \times \frac{1}{2}-(-2) \\ = 1\]

(b)\[ \int_{-1}^{1} (2e^x+\sin{x}) \ dx = (2e^x-\cos{x})\mid_{-1}^{1} \\ = (2e-\cos{1})-(2e^{-1}-\cos{(-1)}) \\ = 2e-\cos{1}-2e^{-1}+\cos{(-1)} \\ = 2e-\cos{1}-2e^{-1}+\cos{1} \\ = 2e-2e^{-1}\\\]

(c)\[ \int_{-4}^{-1} (e^x+\frac{1}{x}) \ dx = (e^x+\ln{\mid x \mid})\mid_{-4}^{-1} \\ = (e^{-1}+ln1)-(e^{-4}+ln4) \\ = e^{-1}-e^{-4}-ln4 \\ \]

(d)\[ \int_{1}^{2} \frac{2-x}{x} \ dx = \int_{1}^{2} (\frac{2}{x}-1) \ dx \\ =(2\ln{\mid x \mid}-x)\mid_{1}^{2} \\ = (2\ln2-2)-(2\ln1-1) \\ = 2\ln2-1 \\ \]

Q3

(a)\[ \int_{0}^{\pi} \sin{x} \ dx = (-\cos{x}) \mid_{0}^{\pi} \\ = (-\cos{\pi})-(-\cos{0}) \\ = 1-(-1) \\ = 2 \]

(b)\[ \int_{\frac{1}{2}}^{e} \frac{1}{x} \ dx = (\ln{\mid x \mid}) \mid_{\frac{1}{2}}^{e} \\ = (\ln{e})-(\ln{\frac{1}{2}}) \\ = 1+\ln{2} \]

(c)\[ \int_{0}^{2} e^{-x} \ dx = -e^{-x} \mid_{0}^{2} \\ = -(e^{-2}-1) \\ = 1-e^{-2}\\ \int_{-1}^{0} e^{x} \ dx = e^{x} \mid_{-1}^{0} \\ = 1-e^{-1} \\ thus\ the\ area\ is: 1-e^{-2}+1-e^{-1}=2-e^{-2}-e^{-1}\]

Exercise 10.2

Q1

(c)\[Put \ u=x^2, \ du=2x\ dx.\\ \int x\sin{x^2} \ dx = \int \frac{1}{2} \sin{x^2} (2x)\ dx \\ = \int \frac{1}{2} \sin{u} \ du \\ = -\frac{1}{2} \cos{u} +C \\ = -\frac{1}{2} \cos{x^2} +C \]

(d)\[Put \ u=\cos{x}, \ du=-\sin{x}\ dx.\\ \int \sin{x} \cos^2{x} \ dx = \int - \cos^2{x} (-\sin{x})\ dx \\ = \int -u^2 \ du \\ = -\frac{1}{3} u^3 +C \\ = -\frac{1}{3} \cos^3{x} +C \]

(f)\[Put \ u=e^x, \ du=e^xdx.\\ \int e^x \sec^2{(e^x)} \ dx = \int \sec^2({e^x}) \ e^xdx \\ = \int \sec^2{u} \ du \\ = \tan{u} +C \\ = \tan{(e^x)} +C \]

(h)\[Put \ u=x^3-1, \ du=3x^2 \ dx.\\ \int x^2e^{x^3-1} \ dx = \int \frac{1}{3}e^{x^3-1} (3x^2)\ dx \\ = \int \frac{1}{3}e^{u} \ du \\ = \frac{1}{3}e^{u} +C \\ = \frac{1}{3}e^{x^3-1} +C \]

(i)\[Put \ u=x^2+1, \ du=2x \ dx.\\ \int \frac{x}{x^2+1} \ dx = \int \frac{1}{2(x^2+1)} (2x)\ dx \\ = \int \frac{1}{2u} \ du \\ = \frac{1}{2}\ln\mid{u}\mid +C \\ = \frac{1}{2}\ln\mid{x^2+1}\mid +C\\ = \frac{1}{2}\ln{(x^2+1)} +C \]

(j)\[Put \ u=\frac{1}{x}, \ du=-\frac{1}{x^2} \ dx.\\ \int \frac{\sin{\frac{1}{x}}}{x^2} \ dx = \int -\sin{\frac{1}{x}} (-\frac{1}{x^2}) \ dx \\ = \int -\sin{u} \ du \\ = \cos{u} +C \\ = \cos{(\frac{1}{x})} +C \]

(k)\[Put \ u=x^2+2x+3, \ du=2x+2 \ dx.\\ \int (x+1)(x^2+2x+3)^7 \ dx = \int \frac{1}{2}(x^2+2x+3)^7 (2x+2) \ dx \\ = \int \frac{1}{2}u^7 \ du \\ = \frac{1}{16}u^8 +C \\ = \frac{1}{16}(x^2+2x+3)^8 +C \]

(m)\[Put \ u=e^x-3x, \ du=(e^x-3) \ dx.\\ \int (e^x-3x)^4(e^x-3) \ dx = \int u^4 \ du \\ = \frac{1}{5}u^5 +C \\ = \frac{1}{5}(e^x-3x)^5 +C \] (n)\[Put \ u=\sqrt{x}, \ du=\frac{1}{2\sqrt{x}}\ dx.\\ \int \frac{e^{\sqrt{x}}}{\sqrt{x}} \ dx = \int 2e^{\sqrt{x}} (\frac{1}{2\sqrt{x}}) \ dx \\ = \int 2e^{u} \ du \\ = 2e^u +C \\ = 2e^{\sqrt{x}}+C \]

(o)\[Put \ u=\ln{(x+1)}, \ du=\frac{1}{x+1}\ dx.\\ \int \frac{\ln{(x+1)}}{x+1} \ dx = \int \ln{(x+1)}(\frac{1}{x+1}) \ dx \\ = \int u \ du \\ = \frac{1}{2}u^2 +C \\ = \frac{1}{2}(\ln{(x+1)})^2+C \\ Note:x+1>0\ as\ there\ is\ \ln(x+1)\ in \ the \ original\ expression. \]

(t)\[Put \ u=x+\frac{1}{x}, \ du=(1-\frac{1}{x^2})\ dx.\\ \int \frac{(x^2-1)e^{x+\frac{1}{x}}}{x^2} \ dx = \int e^{x+\frac{1}{x}}(1-\frac{1}{x^2}) \ dx \\ = \int e^u \ du \\ = e^u +C \\ = e^{x+\frac{1}{x}}+C \]

Q2

(c)\[Put \ u=x^3, \ du=3x^2\ dx;when\ x=0,u=0\ and\ when\ x=1,u=1\\ \int_0^1 x^2 \cos{x^3} \ dx = \int_0^1 \frac{1}{3}\cos{x^3} (3x^2) \ dx \\ = \int_0^1 \frac{1}{3}\cos{u} \ du \\ = \frac{1}{3}\sin{u}\mid_0^1 \\ = \frac{1}{3}(\sin{1}-\sin{0}) \\ = \frac{1}{3}\sin{1} \]

(f)\[Put \ u=\ln{x}, \ du=\frac{1}{x}\ dx;when\ x=1,u=0\ and\ when\ x=e^{\pi},u=\pi \\ \int_1^{e^{\pi}} \frac{\sin{(\ln{x}})}{x} \ dx = \int_1^{e^{\pi}} \sin{(\ln{x}}) \frac{1}{x} \ dx \\ = \int_0^{\pi} \sin{u} \ du \\ = -\cos{u}\mid_0^{\pi} \\ = -(\cos{\pi}-\cos{0}) \\ = -(-1-1) \\ = 2 \]

(h)\[Put \ u=\ln{x}, \ du=\frac{1}{x}\ dx;when\ x=e,u=1\ and\ when\ x=e^2,u=2 \\ \int_e^{e^2} \frac{1}{x\ln{x}} \ dx = \int_e^{e^2} \frac{1}{\ln{x}} \frac{1}{x} \ dx \\ = \int_1^2 \frac{1}{u} \ du \\ = \ln{\mid{u}\mid}\mid_1^2 \\ = \ln{2}-\ln{1} \\ = \ln{2} \]

(j)\[Put \ u=\sqrt{x+1}, \ du=\frac{1}{2\sqrt{x+1}}\ dx,\ x=u^2-1;when\ x=0,u=1\ and\ when\ x=8,u=3 \\ \int_0^8 \frac{x}{\sqrt{x+1}} \ dx = \int_0^8 2x(\frac{1}{2\sqrt{x+1}}) \ dx \\ = \int_1^3 2(u^2-1) \ du \\ = (\frac{2}{3}u^3-2u)\mid_1^3 \\ = (\frac{2}{3}\times3^3-2\times3)-(\frac{2}{3}\times1^3-2\times1) \\ = \frac{40}{3} \]

Q3

(a)\[Put \ u=x^2, \ du=2x\ dx;when\ x=0,u=0\ and\ when\ x=\sqrt{\pi},u=\pi \\ \int_0^{\sqrt{\pi}} x\sin{x^2} \ dx = \int_0^{\sqrt{\pi}} \frac{1}{2}\sin{x^2}(2x) \ dx \\ = \int_0^{\pi} \frac{1}{2}\sin{u} \ du \\ = (-\frac{1}{2}\cos{u})\mid_0^{\pi} \\ = -\frac{1}{2}(\cos{\pi}-\cos{0}) \\ = -\frac{1}{2}(-1-1) \\ = 1\]

(b)\[Put \ u=x^2, \ du=2x\ dx;when\ x=0,u=0\ and\ when\ x=1,u=1 \\ \int_0^1 xe^{x^2} \ dx = \int_0^1 \frac{1}{2}e^{x^2}(2x) \ dx \\ = \int_0^1 \frac{1}{2}e^u \ du \\ = (\frac{1}{2}e^u)\mid_0^1 \\ = \frac{1}{2}(e-1)\\ \int_0^1 x \ dx = \frac{1}{2}x^2\mid_0^1=\frac{1}{2}\\ Thus\ the\ area\ is: \\ \frac{1}{2}(e-1)-\frac{1}{2}=\frac{1}{2}e-1\]

Exercise 10.3

Q1

Evaluate the integral using integration by parts

(a)\[ \int xe^{-2x}\ dx \]

(b)\[ \int \sqrt{x}\ln{x}\ dx \]

(c)\[ \int (\ln{x})^2\ dx \]

(d)\[ \int \tan^{-1}{x}\ dx \]

Q2

First make a substitution and then use integration by parts to evaluate the integral.

(a)\[ \int \cos{(\ln{x})}\ dx \]

(b)\[ \int \frac{arcsin{(\ln{x})}}{x}\ dx \]

Q3

Use the integration by parts to prove the reduction formula.

\[ \int \cos^n{x}\ dx = \frac{1}{n}\cos^{n-1}{x}\sin{x}+\frac{n-1}{n}\int \cos^{n-2}{x}\ dx \]

Solutions to Exercise 10.3

Q1

Evaluate the integral using integration by parts \[ \int u\ dv = uv-\int v \ du \] (a)\[ Put \ u=x, \ dv=e^{-2x}\ dx;\ Then \ du=dx,\ v=-\frac{1}{2}e^{-2x} \\ \int xe^{-2x}\ dx = -\frac{1}{2}xe^{-2x}-(\int -\frac{1}{2}e^{-2x})\ dx \\ = -\frac{1}{2}xe^{-2x}-\frac{1}{4}e^{-2x}+C \]

(b)\[ Put \ u=\ln{x}, \ dv=\sqrt{x}\ dx;\ Then \ du=\frac{1}{x}\ dx,\ v=\frac{2}{3}x^{\frac{3}{2}} \\ \int \sqrt{x}\ln{x}\ dx = \frac{2}{3}x^{\frac{3}{2}}\ln{x}-\int \frac{2}{3}x^{\frac{3}{2}} \times \frac{1}{x}\ dx \\ = \frac{2}{3}x^{\frac{3}{2}}\ln{x}-\int \frac{2}{3}x^{\frac{1}{2}}\ dx \\ = \frac{2}{3}x^{\frac{3}{2}}\ln{x}-\frac{4}{9}x^{\frac{3}{2}}+C \]

(c)\[ Put \ u=(\ln{x})^2, \ dv=dx;\ Then \ du=2\frac{\ln{x}}{x}\ dx,\ v=x \\ \int (\ln{x})^2\ dx = x(\ln{x})^2-\int x\times(2\frac{\ln{x}}{x}) \ dx \\ = x(\ln{x})^2-2\int \ln{x}\ dx \\ As\ for\ \int \ln{x}\ dx \\ Put \ u=\ln{x}, \ dv=dx;\ Then \ du=\frac{1}{x}\ dx,\ v=x \\ \int \ln{x}\ dx = x\ln{x}-\int x\times(\frac{1}{x}) \ dx \\ = x\ln{x}-\int 1\ dx \\ = x\ln{x}-x+C' \\ \int (\ln{x})^2\ dx = x(\ln{x})^2-2(x\ln{x}-x)+C \]

(d)\[ Put \ u=\tan^{-1}{x}, \ dv=dx;\ Then \ du=\frac{1}{x^2+1}\ dx,\ v=x \\ \int \tan^{-1}{x}\ dx = x\tan^{-1}{x}-\int \frac{x}{x^2+1}\ dx \\ As\ for\ \int \frac{x}{x^2+1}\ dx \\ Put \ t=x^2+1, \ dt=2x \ dx \\ \int \frac{x}{x^2+1}\ dx = \frac{1}{2}\int \frac{1}{x^2+1}(2x)\ dx\\ = \frac{1}{2}\int \frac{1}{t}\ dt\\ = \frac{1}{2}\ln{\mid t \mid} +C' \\ = \frac{1}{2}\ln{(x^2+1)} +C' \\ \int \tan^{-1}{x}\ dx = x\tan^{-1}{x}-\frac{1}{2}\ln{(x^2+1)}+C \]

Q2

First make a substitution and then use integration by parts to evaluate the integral.

(a)\[ Put \ u=x, \ dv=\frac{\cos{(\ln{x}})}{x} \ dx;\ Then \ du=dx,\ v=\sin{(\ln{x})} \\ \int \cos{(\ln{x})}\ dx = \int x\frac{\cos{(\ln{x})}}{x}\ dx \\ = x\sin{(\ln{x})}-\int \sin{(\ln{x})}\ dx \\ As\ for\ \int \sin{(\ln{x})}\ dx \\ Put \ u=x, \ dv=-\frac{\sin{(\ln{x}})}{x} \ dx;\ Then \ du=dx,\ v=\cos{(\ln{x})} \\ \int \sin{(\ln{x})}\ dx = -\int x(-\frac{\sin{(\ln{x})}}{x})\ dx \\ = -[x\cos{(\ln{x})}-(\int \cos{(\ln{x})}\ dx)] \\ = -x\cos{(\ln{x})}+\int \cos{(\ln{x})}\ dx \\ \int \cos{(\ln{x})}\ dx = x\sin{(\ln{x})}-\int \sin{(\ln{x})}\ dx \\ = x\sin{(\ln{x})}+x\cos{(\ln{x})}-\int \cos{(\ln{x})}\ dx \\ 2\int \cos{(\ln{x})}\ dx = x\sin{(\ln{x})}+x\cos{(\ln{x})}\\ \int \cos{(\ln{x})}\ dx = \frac{x\sin{(\ln{x})}+x\cos{(\ln{x})}}{2}+C\\ \]

(b)\[ Put \ t=\ln{x}, \ dt=\frac{1}{x} \ dx\\ \int \frac{arcsin{(\ln{x})}}{x}\ dx = \int arcsin{t}\ dt \\ Put \ u=arcsin{t}, \ dv=dt;\ Then \ du=\frac{1}{\sqrt{1-t^2}}\ dt,\ v=t \\ \int arcsin{t}\ dt = t \arcsin{t}-\int \frac{t}{\sqrt{1-t^2}}\ dt \\ As\ for\ \int \frac{t}{\sqrt{1-t^2}}\ dt\\ Put\ z=1-t^2,\ dz= -2t\ dt\\ \int \frac{t}{\sqrt{1-t^2}} dt = -\frac{1}{2}\int \frac{1}{\sqrt{1-t^2}}(-2t)\ dt \\ = -\frac{1}{2}\int \frac{1}{\sqrt{z}}\ dz \\ = -\sqrt{z}+C\\ = -\sqrt{1-t^2}+C\\ \int arcsin{t}\ dt = t \arcsin{t} + \sqrt{1-t^2}+C\\ \int \frac{arcsin{(\ln{x})}}{x}\ dx = \ln{x} \arcsin{(\ln{x})} + \sqrt{1-(\ln{x})^2}+C\\ \]

Q3

Use the integration by parts to prove the reduction formula.

\[ Put \ u=\cos^{n-1}{x}, \ dv=\cos{x}\ dx;\ Then \ du=-(n-1)\cos^{n-2}{x}\sin{x}\ dx,\ v=\sin{x} \\ \int \cos^n{x}\ dx = \cos^{n-1}{x}\sin{x}-\int -(n-1)\cos^{n-2}{x}\sin^2{x}\ dx \\ = \cos^{n-1}{x}\sin{x}+(n-1)\int \cos^{n-2}{x}\sin^2{x}\ dx \\ = \cos^{n-1}{x}\sin{x}+(n-1)[\int \cos^{n-2}{x}(1-\cos^2{x})\ dx] \\ = \cos^{n-1}{x}\sin{x}+(n-1)\int \cos^{n-2}{x}\ dx -(n-1) \int \cos^n{x}\ dx \\ n \int \cos^n{x}\ dx = \cos^{n-1}{x}\sin{x}+(n-1)\int \cos^{n-2}{x}\ dx \\ \int \cos^n{x}\ dx = \frac{1}{n}\cos^{n-1}{x}\sin{x}+\frac{n-1}{n}\int \cos^{n-2}{x}\ dx \\ \]

Exercise 10.4

Q1

Evaluate the integral.

(a)\[ \int \frac{5x+1}{(2x+1)(x-1)}\ dx \]

(b)\[ \int \frac{x^2+2x-1}{2x^3+3x^2-2x} \ dx \]

Q2

Make a substitution to express the integrand as a rational function and then evaluate the integral.

(a)\[ \int \frac{1}{x\sqrt{x-1}}\ dx \]

(b)\[ \int \frac{\sin{x}}{\cos^2{x}-3\cos{x}} \ dx \]

Q3

Use integration by parts and integration of rational function to evaluate the integral.

\[ \int \ln{(x^2-x+2)} \ dx \]

Solutions to Exercise 10.4

Q1

Evaluate the integral.

(a)\[ \int \frac{5x+1}{(2x+1)(x-1)}\ dx = \int (\frac{1}{2x+1}+\frac{2}{x-1})\ dx \\ = \int \frac{1}{2x+1}\ dx +\int \frac{2}{x-1}\ dx \\ = \frac{1}{2}\int \frac{1}{2x+1}\ d(2x+1) +\int \frac{2}{x-1}\ d(x-1) \\ = \frac{1}{2}\int \frac{1}{u}\ du +\int \frac{2}{v}\ dv \\ = \frac{1}{2}\ln{\mid u \mid} + 2\ln{\mid v \mid} +C \\ = \frac{1}{2}\ln{\mid (2x+1) \mid} + 2\ln{\mid (x-1) \mid} +C\]

(b)\[ \int \frac{x^2+2x-1}{2x^3+3x^2-2x} \ dx = \int \frac{x^2+2x-1}{x(x+2)(2x-1)}\ dx \\ = \int (\frac{1}{2x}-\frac{1}{10(x+2)}+\frac{1}{5(2x-1)})\ dx \\ = \int \frac{1}{2x}\ dx-\int \frac{1}{10(x+2)}\ dx+\int\frac{1}{5(2x-1)})\ dx \\ = \frac{1}{2}\ln{\mid x \mid}-\frac{1}{10}\ln{\mid (x+2)\mid}+\frac{1}{10}\ln{\mid (2x-1)\mid} +C \\ \]

Q2

Make a substitution to express the integrand as a rational function and then evaluate the integral.

(a)\[ Put\ t=\sqrt{x-1}, x= t^2+1, dt= \frac{1}{2\sqrt{x-1}}\ dx \\ \int \frac{1}{x\sqrt{x-1}}\ dx = \int \frac{2}{x}\frac{1}{2\sqrt{x-1}}\ dx \\ = \int \frac{2}{t^2+1}\ dt \\ = 2\arctan{t} +C \\ = 2\arctan{(\sqrt{x-1})}+C \\\]

(b)\[ Put\ t=\cos{x}, dt= -\sin{x}\ dx \\ \int \frac{\sin{x}}{\cos^2{x}-3\cos{x}} \ dx = -\int \frac{-\sin{x} \ dx}{\cos^2{x}-3\cos{x}} \\ = -\int \frac{1}{t^2-3t}\ dt \\ = -\int \frac{1}{t(t-3)}\ dt \\ = -\frac{1}{3}\int (\frac{1}{t-3} - \frac{1}{t})\ dt \\ = -\frac{1}{3}\int \frac{1}{t-3}\ dt + \frac{1}{3}\int \frac{1}{t}\ dt \\ = -\frac{1}{3}\ln{\mid (t-3) \mid} +\frac{1}{3}\ln{\mid t \mid}+C \\ = \frac{1}{3}\ln{\mid \cos{x} \mid}-\frac{1}{3}\ln{\mid (\cos{x}-3) \mid}+C \\ \]

7.3.1

Q3 Use integration by parts and integration of rational function to evaluate the integral.

Answer:

\[ Put \ u=\ln{(x^2-x+2)}, \ dv=dx;\ Then \ du=\frac{2x-1}{x^2-x+2}\ dx,\ v=x \\ \int \ln{(x^2-x+2)} \ dx = x\ln{(x^2-x+2)}- \int \frac{x(2x-1)}{x^2-x+2}\ dx \\ As\ for\ \int \frac{x(2x-1)}{x^2-x+2}\ dx \\ \int \frac{x(2x-1)}{x^2-x+2}\ dx = \int \frac{2(x^2-x+2)+x-4}{x^2-x+2}\ dx \\ = \int (2+\frac{x-4}{x^2-x+2})\ dx \\ = \int [2+\frac{2x-1}{2(x^2-x+2)}-\frac{7}{2(x^2-x+2)}]\ dx \\ = \int 2\ dx+\int \frac{2x-1}{2(x^2-x+2)}\ dx - \int \frac{7}{2(x^2-x+2)}\ dx \\ = 2x + \int \frac{1}{2(x^2-x+2)}\ d(x^2-x+2) - \frac{7}{2}\int \frac{1}{(x-\frac{1}{2})^2+\frac{7}{4}}\ dx \\ = 2x+\frac{1}{2}\ln{\mid x^2-x+2 \mid}-\int\frac{7}{2}\frac{1}{(x-\frac{1}{2})^2+ \frac{7}{4}}\ dx\\ As\ for\ \int \frac{1}{(x-\frac{1}{2})^2+ \frac{7}{4}}\ dx\\ \int \frac{1}{x^2+a^2} = \frac{1}{a}arctan{\frac{x}{a}}+C \\ \int \frac{1}{(x-\frac{1}{2})^2+ \frac{7}{4}}\ dx = \frac{2}{\sqrt{7}}\arctan{\frac{x-\frac{1}{2}}{\frac{\sqrt{7}}{2}}}+C\\ = \frac{2}{\sqrt{7}}\arctan{\frac{2x-1}{\sqrt{7}}}+C\\ \int \frac{x(2x-1)}{x^2-x+2}\ dx = 2x+\frac{1}{2}\ln{\mid x^2-x+2 \mid}-\sqrt{7}\arctan{\frac{2x-1}{\sqrt{7}}}+C\\ \int \ln{(x^2-x+2)} \ dx = x\ln{(x^2-x+2)}-2x-\frac{1}{2}\ln{\mid x^2-x+2 \mid}+\sqrt{7}\arctan{\frac{2x-1}{\sqrt{7}}}+C \\ \]

Exercise 10.5

Q1

if a birth rate of a population is \[ b(t)=2200+52.3t+0.74t^2 \] people per year and death rate is \[ d(t)=1460+28.8t\] people per year,find the area between these curves for \[ 0\leq t \leq 10 \]What does this area represent?

Q2

(a)Prove that if f is a continuous function, then \[ \int_0^a f(x)\ dx = \int_0^a f(a-x)\ dx \]

(b)Use part(a) to show that \[ \int_0^{\pi/2} \frac{\sin^{n}{x}}{\sin^{n}{x}+\cos^{n}{x}} dx = \frac{\pi}{4} \] for all positive numbers n.

Q3

if n is a positive integer,prove that \[ \int_0^1 (\ln{x})^n\ dx=(-1)^{n}n! \]

Solutions to Exercise 10.5

Q1

if a birth rate of a population is \[ b(t)=2200+52.3t+0.74t^2 \] people per year and death rate is \[ d(t)=1460+28.8t\] people per year,find the area between these curves for \[ 0\leq t \leq 10 \]What does this area represent?

Answer:

\[ \int_0^{10}[b(t)-d(t)]\ dt = \int_0^{10}[(2200+52.3t+0.74t^2)-(1460+28.8t)]\ dt \\ = \int_0^{10}(740+23.5t+0.74t^2)\ dt \\ = (740t+11.75t^2+\frac{0.74}{3}t^3)\mid_0^{10} \\ \approx 8822 \]

The area represents an increase in the number of people compared to last year.

Q2

(a)Prove that if f is a continuous function, then \[\int_0^a f(x)\ dx = \int_0^a f(a-x)\ dx \]

Answer:

\[ Put\ t=a-x,then\ x=a-t.\ dx=-dt. \ when\ x=0,t=a;\ x=a,t=0.\\ \int_0^a f(x)\ dx = \int_a^0 f(a-t)\ (-dt)\\ = \int_0^a f(a-t)\ dt \\ = \int_0^a f(a-x)\ dx \]

(b)Use part(a) to show that \[ \int_0^{\pi/2} \frac{\sin^{n}{x}}{\sin^{n}{x}+\cos^{n}{x}} dx = \frac{\pi}{4} \] for all positive numbers n.

Answer:

\[ As\ proved\ in\ (a), \int_0^a f(x)\ dx = \int_0^a f(a-x)\ dx \\ When\ a=\frac{\pi}{2},\ f(x)=\frac{\sin^{n}{x}}{\sin^{n}{x}+\cos^{n}{x}},\\ f(a-x)=\frac{\sin^{n}{(\frac{\pi}{2}-x)}}{\sin^{n}{(\frac{\pi}{2}-x)}+\cos^{n}{(\frac{\pi}{2}-x)}}\\ = \frac{\cos^{n}{x}}{\cos^{n}{x}+\sin^{n}{x}}\\ \int_0^{\pi/2} \frac{\sin^{n}{x}}{\sin^{n}{x}+\cos^{n}{x}} dx=\int_0^{\pi/2} \frac{\cos^{n}{x}}{\cos^{n}{x}+\sin^{n}{x}}\ dx \\ 2\int_0^{\pi/2} \frac{\sin^{n}{x}}{\sin^{n}{x}+\cos^{n}{x}} dx=\int_0^{\pi/2} \frac{\sin^{n}{x}}{\sin^{n}{x}+\cos^{n}{x}} dx+\int_0^{\pi/2} \frac{\cos^{n}{x}}{\cos^{n}{x}+\sin^{n}{x}}\ dx \\ =\int_0^{\pi/2} \frac{\sin^{n}{x}+\cos^{n}{x}}{\sin^{n}{x}+\cos^{n}{x}} dx \\ =\int_0^{\pi/2} 1 \ dx\\ = x\mid_0^{\pi/2}\\ =\pi/2\\ \int_0^{\pi/2} \frac{\sin^{n}{x}}{\sin^{n}{x}+\cos^{n}{x}} dx = \frac{\pi}{4} \\\]

Q3

if n is a positive integer,prove that \[ \int_0^1 (\ln{x})^n\ dx=(-1)^{n}n! \]

Answer:

\[ Use\ the\ integration\ by\ parts\ for\ many\ times \\ \int_0^1 (\ln{x})^n\ dx=x(\ln{x})^n\mid_0^1-\int_0^1 x [n(\ln{x})^{n-1}\frac{1}{x}]\ dx \\ = [1(\ln{1})^n-0]-\int_0^1 n(\ln{x})^{n-1}\ dx \\ = 0-n\int_0^1 (\ln{x})^{n-1}\ dx \\ = -nx(\ln{x})^{n-1}\mid_0^1+n \int_0^1 x [(n-1)(\ln{x})^{n-2}\frac{1}{x}]\ dx \\ = 0+n(n-1)\int_0^1 (\ln{x})^{n-2}\ dx \\ = n(n-1)x(\ln{x})^{n-2}\mid_0^1-n(n-1) \int_0^1 x [(n-2)(\ln{x})^{n-3}\frac{1}{x}]\ dx \\ = 0 - n(n-1)(n-2)\int_0^1 (\ln{x})^{n-3}\ dx \\ = .....\\ = (-1)^{n}n!\int_0^1 (\ln{x})^0\ dx \\ = (-1)^{n}n!\int_0^1 1\ dx \\ = (-1)^{n}n!x\mid_0^1 \\ = (-1)^{n}n! \]