Tutorial 8: Integration

Tutorial Question

Tutorial Solution

1. Find

   $$\int \ln \left(x^{2}+2\right) d x$$
   Solution

   Let $u=\ln\left( x^{2} +2\right)$, $dv=dx$

   $$\begin{aligned} u & =\ln\left( x^{2} +2\right) & dv & =dx \\ du & =\frac{2x}{x^{2} +2} dx & v &=x \end{aligned}$$

   So,

   $$\begin{aligned} \int \ln\left( x^{2} +2\right) dx & =x\ln\left( x^{2} +2\right) -2\int \frac{x^{2}}{x^{2} +2} dx\\ & =x\ln\left( x^{2} +2\right) -2\int \left( 1-\frac{2}{x^{2} +2}\right) dx\\ & =x\ln\left( x^{2} +2\right) -2x+\frac{4}{\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right) +C\\ & =x\left(\ln\left( x^{2} +2\right) -2\right) +2\sqrt{2}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right) +C\end{aligned}$$

2. Find

   $$\int x^{2} \ln x d x$$
   Solution

   Let $u=\ln x$, $dv=x^{2} \ dx$

   $$\begin{aligned} u & =\ln x &dv & =x^{2} \ dx \\ du & =\frac{dx}{x} &v & =\frac{x^{3}}{3} \end{aligned}$$

   So,

   $$\begin{aligned} \int x^{2}\ln x\ dx & =\frac{x^{3}}{3}\ln x-\int \frac{x^{3}}{3}\frac{dx}{x}\\ & =\frac{x^{3}}{3}\ln x-\frac{1}{3}\int x^{2} dx\\ & =\frac{x^{3}}{3}\ln x-\frac{1}{9} x^{3} +C \end{aligned}$$

3. Find

   $$\int x^{3} e^{x^{2}} d x$$
   Solution

   Let $u=x^{2}$ and $dv=xe^{x^{2}} dx$

   $$\begin{aligned} u & =x^{2} &dv &=xe^{x^{2}} \\ du & =2x\ dx &v & =\frac{1}{2} e^{x^2} \end{aligned}$$

   Hence,

   $$\begin{aligned} \int x^{3} e^{x^{2}} \ dx & =\frac{1}{2} x^{2} e^{x^{2}} -\int xe^{x^{2}} dx\\ & =\frac{1}{2} x^{2} e^{x^{2}} -\frac{1}{2} e^{x^{2}} +C\\ & =\frac{1}{2} e^{x^{2}}\left( x^{2} -1\right) +C \end{aligned}$$

4. Find

   $$\int \frac{(x+1)}{x^{3}+x^{2}-6 x} d x$$
   Solution

   Factoring the denomenator $x^{3} +x^{2} -6x=x\left( x^{2} +x-6\right) =x( x-2)(x+3)$. The integrand is now $\frac{( x+1)}{x(x-2)(x+3)}$.

   Representing the integrand such that:

   $$\frac{( x+1)}{x( x-2)( x+3)} =\frac{A}{x} +\frac{B}{x-2} +\frac{C}{x+3} \tag{1}$$

   Multiplying equation (1) with $x( x-2)( x+3)$,

   $$x+1=A( x-2)( x+3) +Bx( x+3) +Cx( x-2) \tag{2}$$

   Let $x=0$ in (2), $1=A( -2)( 3) +B( 0)( 3) +C( 0)( -2) \ \Rightarrow \ 1=-6A$. So, $A=-\frac{1}{6}$.

   Let $x=2$ in (2), $3=A( 0)( 5) +B( 2)( 5) +C( 2)( 0) \ \ \Rightarrow \ 3=10B$. So, $B=\frac{3}{10}$.

   Let $x=-3$ in (2), $-2=A( -5)( 0) +B( -3)( 0) +C( -3)( -5) \ \Rightarrow \ -2=15C$. So, $C=-\frac{2}{15}$.

   Therefore,

   $$\begin{aligned} \int \frac{( x+1)}{x^{3} +x^{2} -6x} \ dx & =\int \left( -\frac{1}{6}\frac{1}{x} +\frac{3}{10}\frac{1}{x-2} -\frac{2}{15}\frac{1}{x+3}\right) \ dx\\ & =-\frac{1}{6}\ln |x|\ +\frac{3}{10}\ln |x-2|-\frac{2}{15}\ln |x+3|\ +C \end{aligned}$$

5. Find

   $$\int \frac{x^{3}+x^{2}+x+2}{x^{4}+3 x^{2}+2} d x$$
   Solution

   Factoring the denomenator $x^{4} +3x^{2} +2=\left( x^{2} +1\right)\left( x^{2} +2\right)$.

   The integrand is now $\displaystyle\frac{x^{3} +x^{2} +x+2}{\left( x^{2} +1\right)\left( x^{2} +2\right)}$.

   Representing the integrand such that:

   $$\frac{x^{3} +x^{2} +x+2}{\left( x^{2} +1\right)\left( x^{2} +2\right)} =\frac{Ax+B}{x^{2} +1} +\frac{Cx+D}{x^{2} +2} \tag{3}$$

   Multiplying equation (3) with $\left( x^{2} +1\right)\left( x^{2} +2\right)$,

   $$\begin{aligned}[b] x^{3} +x^{2} +x+2 & =( Ax+B)\left( x^{2} +2\right) +( Cx+D)\left( x^{2} +1\right)\\ & =Ax^{3} +2Ax+Bx^{2} +2B+Cx^{3} +Cx+Dx^{2} +D\\ & =( A+C) x^{3} +( B+D) x^{2} +( 2A+C) x+( 2B+D) \end{aligned} \tag{4}$$

   Comparing LHS and RHS of equation (4),

   $$\begin{aligned} A+C & =1\\ B+D & =1\\ 2A+C & =1\\ 2B+D & =2 \end{aligned} \tag{5}$$

   Solving equation (5) simultaneously to obtain $A=0$, $B=1$, $C=1$, $D=0$

   Therefore,

   $$\begin{aligned} \int \frac{x^{3} +x^{2} +x+2}{\left( x^{2} +1\right)\left( x^{2} +2\right)} dx & =\int \left(\frac{1}{x^{2} +1} +\frac{x}{x^{2} +2}\right) dx\\ & =\tan^{-1} x+\frac{1}{2}\ln\left( x^{2} +2\right) +C \end{aligned}$$

6. Find

   $$\int \tan ^{3}(3 x) \sec ^{4}(3 x) d x$$
   Solution

   $$\begin{aligned} \int \tan^{3}( 3x) \ \sec^{4}( 3x) \ dx & =\int \tan^{3}( 3x) \ \left( 1+\tan^{2}( 3x)\right)\sec^{2}( 3x) \ dx\\ & =\int \tan^{3}( 3x)\sec^{2}( 3x) \ dx+\int \tan^{5}( 3x)\sec^{2}( 3x) \ dx\\ & =\frac{1}{3}\frac{1}{4}\tan^{4}( 3x) +\frac{1}{3}\frac{1}{6}\tan^{6}( 3x) +C\\ & =\frac{1}{12}\tan^{4}( 3x) +\frac{1}{18}\tan^{6}( 3x) +C \end{aligned}$$

7. Find

   $$\int \sin ^{4}(x) \cos ^{7}(x) d x$$
   Solution

   Using trigonometry identity, $\sin^{2} x+\cos^{2} x=1\ \Longrightarrow \ \cos^{2} x=1-\sin^{2} x$

   $$\begin{aligned} \int \sin^{4}( x)\cos^{7}( x) \ dx & =\int \sin^{4}( x)\cos^{6}( x)\cos( x) \ dx\\ & =\int \sin^{4}( x)\left( 1-\sin^{2}( x)\right)^{3}\cos( x) \ dx \end{aligned}$$

   Let $u=\sin x$, $du=\cos( x) \ dx$,

   $$\begin{aligned} \int \sin^{4}( x)\cos^{7}( x) \ dx & =\int u^{4}\left( 1-u^{2}\right)^{3} \ \ du\\ & =\int u^{4}\left( 1-u^{2}\right)\left( 1-u^{2}\right)^{2} du\\ & =\int u^{4}\left( 1-u^{2}\right)\left( 1-2u^{2} +u^{4}\right) du\\ & =\int u^{4}\left( 1-2u^{2} +u^{4} -u^{2} +2u^{4} -u^{6}\right) du\\ & =\int u^{4}\left( 1-3u^{2} +3u^{4} -u^{6}\right) du\\ & =\int u^{4} -3u^{6} +3u^{8} -u^{10} du\\ & =\frac{1}{5} u^{5} -\frac{3}{7} u^{7} +\frac{3}{9} u^{9} -\frac{1}{11} u^{11} +C \end{aligned}$$

   Substituting back $u=\sin x$,

   $$\int \sin^{4}( x)\cos^{7}( x) \ dx=\frac{1}{5}\sin^{5} x-\frac{3}{7}\sin^{7} x+\frac{1}{3}\sin^{9} x-\frac{1}{11}\sin^{11} x+C$$

8. Find

   $$\int \frac{d x}{x^{2} \sqrt{9-x^{2}}}$$
   Solution

   Let $x=3\sin \theta$, therefore $\theta =\sin^{-1}\frac{x}{3}$. Then, $dx=3\cos \theta \ d\theta$, and

   $$\begin{aligned} \sqrt{9-x^{2}} & =\sqrt{9-( 3\sin \theta )^{2}}\\ & =\sqrt{9-9\sin^{2} \theta }\\ & =3\sqrt{1-\sin^{2} \theta }\\ & =3\sqrt{\cos^{2} \theta }\\ & =3\cos \theta \end{aligned}$$

   Using defination of inverse sine, $-\frac{\pi }{2} < \theta < \frac{\pi }{2}$. Therefore, $\cos \theta >0$.

   Thus, $\displaystyle\cos \theta =\vert \cos \theta \vert =\frac{\sqrt{9-x^{2}}}{3}$.

   Hence,

   $$\begin{aligned} \int \frac{dx}{x^{2}\sqrt{9-x^{2}}} \ & =\int \frac{3\cos \theta \ d\theta }{9\sin^{2} \theta ( 3\cos \theta )}\\ & =\frac{1}{9}\int \csc^{2} \theta \ d\theta \\ & =-\frac{1}{9}\cot \theta +C\\ & =-\frac{1}{9}\frac{\cos \theta }{\sin \theta } +C\\ & =-\frac{1}{9}\frac{\frac{\sqrt{9-x^{2}}}{3}}{\frac{x}{3}} +C\\ & =-\frac{1}{9}\frac{\sqrt{9-x^{2}}}{x} +C \end{aligned}$$

Additional exercises

Please integrate

1. $\displaystyle\int \frac{x+7}{x^{2}(x+2)} d x$

   Solution

   Decompose into partial fractions (There is a repeated linear factor!), getting

   $$\int \frac{x+7}{x^2(x+2)} d x=\int\left(\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+2}\right) d x$$

   After getting a common denominator, adding fractions, and equating numerators, it follows that $A x(x+2)+B(x+2)+C x^2=x+7$;

   let $x=0: A(0)+B(2)+C(0)=7 \longrightarrow B=\frac{7}{2}$;

   let $x=-2: A(0)+B(0)+C(4)=5 \longrightarrow C=\frac{5}{4}$;

   let $x=-1: A(-1)+B(1)+C(1)=-A+\frac{7}{2}+\frac{5}{4}=6 \longrightarrow A=-\frac{5}{4}$.

   $$\begin{aligned} \int \frac{x+7}{x^{2}(x+2)} d x&=\int\left(\frac{-5 / 4}{x}+\frac{7 / 2}{x^2}+\frac{5 / 4}{x+2}\right) d x \\ &=\int\left(-(5 / 4) \frac{1}{x}+(7 / 2) x^{-2}+(5 / 4) \frac{1}{x+2}\right) d x \\ &=-(5 / 4) \ln |x|+(7 / 2) \frac{x^{-1}}{(-1)}+(5 / 4) \ln |x+2|+C \\ &=-\frac{5}{4} \ln |x|-\frac{7}{2 x}+\frac{5}{4} \ln |x+2|+C \end{aligned}$$

2. $\displaystyle\int \frac{\cos x}{\sin ^{3} x+\sin x} d x$

   Solution

   Use the method of u-substitution first. Let $u=\sin x$ so that $d u=\cos x d x$.

   Substitute into the original problem, replacing all forms of $x$, getting

   $$\begin{aligned} \int \frac{\cos x}{\sin ^3 x+\sin x} d x&=\int \frac{1}{\sin ^3 x+\sin x} \cos x d x \\ &=\int \frac{1}{u^3+u} d u \end{aligned}$$

   Factor and decompose into partial fractions.

   $$\begin{aligned} &=\int \frac{1}{u\left(u^2+1\right)} d u \\ &=\int\left(\frac{A}{u}+\frac{B u+C}{u^2+1}\right) d u \end{aligned}$$

   After getting a common denominator, adding fractions, and equating numerators, it follows that $A\left(u^2+1\right)+(B u+C) u=1$;

   let $u=0: A(1)+0=1 \longrightarrow A=1$;

   let $u=i: A\left(i^2+1\right)+(B i+C) i=A(0)+B i^2+C i=-B+C i=1$;

   it follows that $B=-1$ and $C=0$.

   $$\begin{aligned} &=\int\left(\frac{1}{u}+\frac{-u}{u^2+1}\right) d u \\ &=\ln |u|-\frac{1}{2} \ln \left|u^2+1\right|+C \\ &=\ln |\sin x|-\frac{1}{2} \ln \left|\sin ^2 x+1\right|+C \end{aligned}$$