Basic Examples On Integration Set-1

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A lot many functions that we’ll encounter can be reduced to simpler forms by some rearrangement/algebraic manipulation. These simpler forms are easily integrable because they correspond to one of the standard basic integrals that we discussed in the previous section.

The rearrangement technique is best illustrated through examples.

Example - 1

Evaluate  \(\begin{align}\int {\frac{{\cos x - \cos 2x}}{{1 - \cos x}}dx} \end{align}\) .

Solution: We can try expanding \(\text{cos }2x\) by the half angle formula:

\[\begin{align}& \ \int{\frac{\cos x-\cos 2x}{1-\cos x}dx}=\int{\frac{\cos x-\left( 2{{\cos }^{2}}x-1 \right)}{1-\cos x}dx} \\  & \qquad \qquad \qquad \qquad\quad=\int{\frac{\left( 2\cos x+1 \right)\left( 1-\cos x \right)}{\left( 1-\cos x \right)}dx} \\ & \qquad \qquad \qquad\qquad\quad=\int{\left( 2\cos x+1 \right)dx} \\  & \qquad \qquad \qquad \qquad\quad=2\sin x+x+C. \\ \end{align}\]

Observe how the rearrangement we used led to a simpler expression that was easily integrable.

Example - 2 

Evaluate  \(\begin{align}\int {\frac{1}{{\sin \left( {x - a} \right)\cos \left( {x - b} \right)}}dx}\end{align} \).

Solution: The denominator is of the form sin P cos Q, where P and Q are variable; but notice an important fact: P – Q is a constant. This should give us the required hint:

\[\begin{align}& \ \int{\frac{1}{\sin \left( x-a \right)\cos \left( x-b \right)}dx=\frac{1}{\cos \left( a-b \right)}\int{\frac{\cos \left( a-b \right)}{\sin \left( x-a \right)\cos \left( x-b \right)}}dx} \\ &  \\  & \qquad \qquad \qquad \qquad \qquad \qquad =\frac{1}{\cos \left( a-b \right)}\int{\frac{\cos \left\{ \left( x-b \right)-\left( x-a \right) \right\}}{\sin \left( x-a \right)\cos \left( x-b \right)}dx} \\  &  \\  & \qquad \qquad \qquad \qquad \qquad \qquad =\frac{1}{\cos \left( a-b \right)}\int{\frac{\cos \left( x-b \right)\cos \left( x-a \right)+\sin \left( x-b \right)\sin \left( x-a \right)}{\sin \left( x-a \right)\cos \left( x-b \right)}dx} \\  &  \\  & \qquad \qquad \qquad \qquad \qquad \qquad =\frac{1}{\cos \left( a-b \right)}\int{\left\{ \cot \left( x-a \right)+\tan \left( x-b \right) \right\}dx} \\  &  \\  & \qquad \qquad \qquad \qquad \qquad \qquad =\frac{1}{\cos \left( a-b \right)}\left\{ \ln \left| \sin \left( x-a \right) \right|-\ln \left| \cos \left( x-b \right) \right| \right\}+C \\ 
 &  \\  & \qquad \qquad \qquad \qquad \qquad \qquad =\frac{1}{\cos \left( a-b \right)}\ln \left| \frac{\sin \left( x-a \right)}{\cos \left( x-b \right)} \right|+C \\ \end{align}\]

What we had to do in this question was therefore to realise that since P – Q is a constant, an introduction of the term cos (P – Q) in the numerator would lead to cancellations and simple ‘cot’ and ‘tan’ terms which can easily be integrated.

Example – 3

Evaluate \(\begin{align}\int {\frac{{{x^3}}}{{{{\left( {x + 1} \right)}^2}}}dx} \end{align}\).

Solution: The numerator has a degree higher then the denominator which hints that some reduction of this rational expression is possible. This reduction can be accomplished if we somehow rearrange the numerator in such a way that it leads to a cancellation of common factors with the denominator; since the denominator is \({\left( {{\rm{ }}x{\rm{ }} + {\rm{ }}1} \right)^2}\) , we try to rearrange the numerator in terms of \({\left( {{\rm{ }}x{\rm{ }} + {\rm{ }}1} \right)}\):

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

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Indefinite Integration
grade 11 | Questions Set 1
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Download practice questions along with solutions for FREE:
Indefinite Integration
grade 11 | Questions Set 1
Indefinite Integration
grade 11 | Answers Set 1
Indefinite Integration
grade 11 | Questions Set 2
Indefinite Integration
grade 11 | Answers Set 2
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