Tutorial 8: Integration

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Tutorial PDF

Class Recording

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Q1: Find \(\int \ln\left( x^{2} +2\right) dx\).

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Solution

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

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Q2: Find \(\int x^{2}\ln x\ dx\).

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Solution

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

So,

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

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Q3: Find \(\int x^{3} e^{x^{2}} dx\).

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Solution

Let \(u=x^{2}\) and \(dv=xe^{x^{2}} dx\).

So,

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

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Q4: Find \(\int \frac{( x+1)}{x^{3} +x^{2} -6x} dx\).

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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 \(x( x-2)( x+3)\),

\[x+1=A( x-2)( x+3) +Bx( x+3) +Cx( x-2)\]

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,

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

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Q5: Find \(\int \frac{x^{3} +x^{2} +x+2}{x^{4} +3x^{2} +2} dx\).

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Solution

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

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

Representing the integrand such that:

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

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

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

Comparing LHS and RHS of equation (2),

\[\begin{aligned} A+C & =1\\ B+D & =1\\ 2A+C & =1\\ 2B+D & =2 \end{aligned}\]

Solving equation (3) simultaneously to obtain \(A=0\), \(B=1\), \(C=1\), \(D=0\)

Therefore,

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

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Q6: Find \(\int \tan^{3}( 3x) \ \sec^{4}( 3x) \ dx\).

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Solution

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

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Q7: Find \(\int \sin^{4}( x)\cos^{7}( x) \ dx\).

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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 \(u=\sin x\),

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

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Q8: Find \(\int \frac{dx}{x^{2}\sqrt{9-x^{2}}}\).

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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, \(\cos \theta =\vert \cos \theta \vert =\frac{\sqrt{9-x^{2}}}{3}\).

Hence,

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

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