MATH101 Calculus Workshop Exercise 5 Solutions

W5.1. (i) The indefinite integral is

$$\int\bigl(5x^{4}+2x^{-1/2}+{{4}\over{x}}-{{6}\over{x^{4}}}\bigr)dx=x^{5}+4x^{1/2}+4\log|x|+2x^{-3}+C.$$

(ii) Likewise we integrate each summand and obtain

$$\int\bigl(e^{1-3x}+\sqrt{5x+7}\bigr)dx={{-1}\over{3}}e^{1-3x}+{{2}\over{15}}(5x+7)^{3/2}+C.$$

W5.2. The definite integrals are

$$(i)\quad\int_{0}^{1}(1-x)^{5/2}\,dx=\Bigl[-{{2}\over{7}}(1-x)^{7/2}\Bigr]_{0}^{1}={{2}\over{7}};$$

$$\eqalignno{(ii)\quad\int_{\pi/4}^{\pi/3}{\hbox{cosec}}^{2}x\,dx&=\Bigl[-\cot x\Bigr]_{\pi/4}^{\pi/3}\cr&=\cot\pi/4-\cot\pi/3\cr&=1-1/\sqrt{3},\cr}$$

where the integral (ii) is similar to example in lectures.

W5.3. (i) We have

$$\eqalignno{{{d}\over{dx}}\bigl(\sec x+\tan x\bigr)&={{d}\over{dx}}\Bigl({{1}\over{\cos x}}+{{\sin x}\over{\cos x}}\Bigr)\cr&={{\sin x}\over{\cos^{2}x}}+{{1}\over{\cos^{2}x}}\cr&=\sec x\tan x+\sec^{2}x;\cr}$$

hence we have

$$\eqalignno{{{d}\over{dx}}\log\bigl|\sec x+\tan x\bigr|&={{\sec x\tan x+\sec^{2}x}\over{\sec x+\tan x}}\cr&=\sec x.\cr}$$

(ii) By the Fundamental Theorem of Calculus and (i), we have

$$\int\sec x\,dx=\log\bigl|\sec x+\tan x\bigr|+C.$$

W5.4. (i) We have

$$\eqalignno{\int_{0}^{\pi}\cos^{2}t\,dt&={{1}\over{2}}\int_{0}^{\pi}(1+\cos 2t)\,dt\cr&={{1}\over{2}}\bigl[t+{{1}\over{2}}\sin 2t\bigr]_{0}^{\pi}\cr&={{1}\over{2}}\bigl[\pi+{{1}\over{2}}\sin 2\pi\bigr]-{{1}\over{2}}\bigl[0\bigr]\cr&={{\pi}\over{2}}.\cr}$$

(ii) We use the addition formulæ and obtain

$$\eqalignno{\int_{0}^{\pi/2}\cos t\cos 2t\,dt&={{1}\over{2}}\int_{0}^{\pi/2}(\cos t+\cos 3t)\,dt\cr&={{1}\over{2}}\bigl[\sin t+{{1}\over{3}}\sin 3t\bigr]_{0}^{\pi/2}\cr&={{1}\over{2}}\bigl[\sin{{\pi}\over{2}}+{{1}\over{3}}\sin{{3\pi}\over{2}}\bigr]-{{1}\over{2}}\bigl[0\bigr]\cr&={{1}\over{2}}\bigl[1-{{1}\over{3}}\bigr]\cr&={{1}\over{3}}.\cr}$$

W5.5 (i) Integrating twice by parts, we have

$$\eqalignno{I&=\int x^{2}\sin x\,dx\cr&=-x^{2}\cos x+\int 2x\cos x\,dx\cr&=-x^{2}\cos x+2x\sin x-\int 2\sin x\,dx\cr&=-x^{2}\cos x+2x\sin x+2\cos x+C.\cr}$$

(ii) Likewise, we have

$$\eqalignno{J&=\int x^{2}\log x\,dx\cr&=(1/3)x^{3}\log x-\int(1/3)x^{3}(1/x)\,dx\cr&=(1/3)x^{3}\log x-\int(1/3)x^{2}\,dx\cr&=(1/3)x^{3}\log x-(1/9)x^{3}+C.\cr}$$

W5.6 (i) Integrating twice by parts, we have

$$\eqalignno{I&=\int x^{2}\cosh x\,dx\cr&=x^{2}\sinh x-\int 2x\sinh x\,dx\cr&=x^{2}\sinh x-2x\cosh x+\int 2\cosh x\,dx\cr&=x^{2}\sinh x-2x\cosh x+2\sinh x+C.\cr}$$

(ii) Using a hyperbolic identity, we obtain

$$\eqalignno{\int\cosh^{2}x\,dx&={{1}\over{2}}\int(\cosh 2x+1)\,dx\cr&={{1}\over{4}}\sinh 2x+{{x}\over{2}}+C.\cr}$$

W5.7 Let

$$I_{n}=\int_{0}^{\pi/2}\sin^{n}xdx.$$

(i) In particular we have

$$I_{0}=\int_{0}^{\pi/2}\sin^{0}x\,dx=\int_{0}^{\pi/2}1dx={{\pi}\over{2}},$$

$$I_{1}=\int_{0}^{\pi/2}\sin x\,dx=\bigl[-\cos x\bigr]_{0}^{\pi/2}=\cos 0-\cos{\pi/2}=1.$$

(ii) We integrate by parts to get, for $n>1$,

$$\eqalign{I_{n}&=\int_{0}^{\pi/2}\sin x\sin^{n-1}x\,dx\cr&=\bigl[-\cos x\sin^{n-1}x\bigr]_{0}^{\pi/2}+(n-1)\int_{0}^{\pi/2}\cos^{2}x\sin^{n-2}x\,dx\cr}$$

since ${{d}\over{dx}}\sin^{n-1}x=(n-1)\sin^{n-2}x\cos x.$ We observe that $\sin^{n-1}0=0$ for $n>1$, and that $\cos^{2}x=1-\sin^{2}x;$ hence

$$\eqalign{I_{n}&=(n-1)\int_{0}^{\pi/2}\sin^{n-2}x\,dx-(n-1)\int_{0}^{\pi/2}\sin^{n}x\,dx\cr&=(n-1)I_{n-2}-(n-1)I_{n},\cr}$$

and we rearrange this to obtain the reduction formula

$$I_{n}={{n-1}\over{n}}I_{n}.$$

By this reduction formula and the results of (i), we have

$$\eqalign{I_{4}&={{3}\over{4}}I_{2}={{3.1}\over{4.2}}I_{0}={{3.1}\over{4.2}}{{\pi}\over{2}}={{3\pi}\over{16}},\cr I_{5}&={{4}\over{5}}I_{3}={{4.2}\over{5.3}}I_{1}={{8}\over{15}}.\cr}$$

please try to write this nicely

W5.8. (i) Now

$${{d}\over{dx}}\sin^{n}x=n\sin^{n-1}x\cos x,$$

so

$$\eqalign{{{d^{2}}\over{dx^{2}}}\sin^{n}x&=n(n-1)\sin^{n-2}x\cos^{2}x-n\sin^{n}x\cr&=n(n-1)\sin^{n-2}x(1-\sin^{2}x)-n\sin^{n}x\cr&=n(n-1)\sin^{n-2}x-n^{2}\sin^{n}x.\cr}$$

(ii) We integrate by parts and obtain

$$\eqalign{I_{n}&=\int_{0}^{\pi}e^{x}\sin^{n}x\,dx\cr&=\bigl[e^{x}\sin^{n}x\bigr]_{0}^{\pi}-n\int_{0}^{\pi}e^{x}\sin^{n-1}x\cos x\,dx.\cr}$$

so, when we integrate by parts again, we obtain

$$\eqalign{I_{n}&=\bigl[e^{x}\sin^{n}x-ne^{x}\sin^{n-1}x\cos x\bigr]_{0}^{\pi}+n\int_{0}^{\pi}e^{x}\bigl((n-1)\sin^{n-2}x-n\sin^{n}x\bigr)dx\cr I_{n}&=n(n-1)I_{n-2}-n^{2}I_{n};\cr}$$

hence we have the recurrence relation

$$I_{n}={{n(n-1)}\over{n^{2}+1}}I_{n-2}\qquad(n>1).$$

Starting from the result for $I_{1}$ in lectures, we deduce

$$\eqalign{I_{1}&={{1}\over{2}}{(e^{\pi}+1)},\cr I_{3}&={{3.2}\over{9+1}}I_{1}={{3}\over{10}}({{e^{\pi}+1}}),\cr I_{5}&={{5.4}\over{25+1}}I_{3}={{3}\over{13}}(e^{\pi}+1),\cr I_{7}&={{7.6}\over{49+1}}I_{5}={{63}\over{325}}(e^{\pi}+1).\cr}$$

W5.9. (i) Observe that $(1+x^{2})^{\prime}=2x$, hence

$$\int{{2x}\over{1+x^{2}}}dx=\log(1+x^{2})+C.$$

Alternatively, let $u=1+x^{2}$, so ${{du}\over{dx}}=2x$, and substitute to get

$$\eqalign{\int{{2xdx}\over{1+x^{2}}}&=\int{{du}\over{u}}\cr&=\log|u|+C\cr&=\log(1+x^{2})+C.\cr}$$

(ii) Observe that ${{d}\over{dx}}(x^{2}+6x+100)=2x+6$, so we let $u=x^{2}+6x+100$ and then $du/dx=2x+6$, so

$$\eqalignno{\int{{(x+3)dx}\over{x^{2}+6x+100}}&={{1}\over{2}}\int{{du}\over{u}}\cr&={{1}\over{2}}\log|u|+C\cr&={{1}\over{2}}\log(x^{2}+6x+100)+C.\cr}$$

W5.10. (i) By partial fractions, we have

$$\eqalign{\int{{2x+3}\over{(x+1)(x+2)}}dx&=\int\Bigl({{1}\over{x+1}}+{{1}\over{x+2}}\Bigr)dx\cr&=\log|x+1|+\log|x+2|+C.\cr}$$

(ii) The partial fractions here give

$$\eqalign{\int{{x}\over{(x-1)^{2}}}&=\int{{x-1+1}\over{(x-1)^{2}}}dx\cr&=\int{{dx}\over{x-1}}dx+\int{{dx}\over{(x-1)^{2}}}\cr&=\log|x-1|-{{1}\over{x-1}}+C.\cr}$$

W5.11. Let $u=\sin^{2}t$ so that ${{du}\over{dt}}=2\cos t\sin t$ and $1-u=1-\sin^{2}t=\cos^{2}t.$ Then

$$\matrix{u&|&0&1\cr t&|&0&\pi/2\cr}$$

and so we obtain by substituting

$$\eqalign{J_{k}&=\int_{0}^{1}(1-u)^{k/2}u^{-1/2}\,du\cr&=2\int_{0}^{\pi/2}\cos^{k}t(\sin t)^{-1}\cos t\sin t\,dt\cr&=2\int_{0}^{\pi/2}\cos^{k+1}t\,dt=2I_{k+1},\cr}$$

where we recall the notation of lectures. In particular,

$$J_{3}=2I_{4}={{3\pi}\over{8}}.$$

W5.12. (i) Let $x=3\tan u$, so that ${{dx}\over{du}}=3\sec^{2}u$ and $9+x^{2}=9(1+\tan^{2}u)=9\sec^{2}u$. The limits of integration change as

$$\eqalign{x&|\quad\sqrt{3}\qquad\rightarrow\infty\cr u&|\quad\pi/6\qquad(\pi/2)-,\cr}$$

Hence

$$\eqalign{\int_{\sqrt{3}}^{\infty}{{dx}\over{9+x^{2}}}&=\int_{\pi/6}^{\pi/2-}{{3\sec^{2}u}\over{3^{2}\sec^{2}u}}du\cr&={{1}\over{3}}\int_{\pi/6}^{\pi/2-}du\cr&=\pi/9.\cr}$$

(ii) In this integral we let $x=3\tan u$, so that ${{dx}\over{du}}=3\sec^{2}u$ and $9+x^{2}=9(1+\tan^{2}u)=9\sec^{2}u$ as before. The limits of integration change as

$$\eqalign{x&|\quad 0\qquad\rightarrow\infty\cr u&|\quad 0\qquad(\pi/2)-,\cr}$$

Hence

$$\eqalign{\int_{0}^{\infty}{{dx}\over{(9+x^{2})^{2}}}&=\int_{0}^{\pi/2-}{{3\sec^{2}u}\over{9^{2}\sec^{4}u}}du\cr&={{1}\over{27}}\int_{0}^{\pi/2-}\cos^{2}u\,du\cr&={{1}\over{2\times 27}}\int_{0}^{\pi/2-}(1+\cos 2u)du\cr&={{1}\over{2\times 27}}\Bigl[u+{{1}\over{2}}\sin 2u\Bigr]_{0}^{\pi/2-}\cr&={{1}\over{2\times 27}}\Bigl[{{\pi}\over{2}}+{{1}\over{2}}\sin\pi\Bigr]-{{1}\over{2\times 27}}\Bigl[0\Bigr]\cr&={{\pi}\over{108}}.}$$

W5.13. (i) We make the substitution $u=x^{2}+4x+5$, so $du/dx=2x+4$ and find the integral

$$\eqalignno{J_{1}&=\int{{2x+4}\over{x^{2}+4x+5}}dx\cr&=\int{{du}\over{u}}\cr&=\log|u|+C\cr&=\log(x^{2}+4x+5)+C.\cr}$$

(ii) In the integral

$$J_{2}=\int{{dx}\over{(x+2)^{2}+1}}$$

we substitute $x+2=\tan t$, so $dx/dt=\sec^{2}t$ and $(x+2)^{2}+1=\tan^{2}t+1=\sec^{2}t$; hence

$$\eqalignno{J_{2}&=\int{{dx}\over{(x+2)^{2}+1}}\cr&=\int{{\sec^{2}tdt}\over{\sec^{2}t}}\cr&=\int dt\cr&=t+C\cr&=\tan^{-1}(x+2)+C.\cr}$$

(iii) Now we look for constants $A$ and $B$ such that

$$\int{{xdx}\over{x^{2}+4x+5}}=A\int{{2x+4}\over{x^{2}+4x+5}}+B\int{{dx}\over{x^{2}+4x+5}}$$

and then

$$x=A(2x+4)+B$$

so $A=1/2$ and $B=-2$; hence

$$\int{{xdx}\over{x^{2}+4x+5}}={{1}\over{2}}\log(x^{2}+4x+5)-2\tan^{-1}(x+2)+C.$$

W5.14. The difference of squares identity gives $(1-x)(1+x)=1-x^{2}$, so

$$\eqalignno{\int{{\sqrt{1+x}}\over{\sqrt{1-x}}}dx&=\int{{1+x}\over{\sqrt{1-x^{2}}}}dx\cr&=\int{{dx}\over{\sqrt{1-x^{2}}}}+\int{{xdx}\over{\sqrt{1-x^{2}}}}\cr&=\sin^{-1}x-\sqrt{1-x^{2}}+C,\cr}$$

where the final integral is given by guesswork or by using the substitution $u=1-x^{2}$.