In the previous section, we completed a discussion on the method of integration by parts. In this section, we will see three standard integrals.
I. The integral $\bf{\int{\left[\sqrt{x^2 - a^2} \right]dx}}$
This can be calculated as follows:
1. Assigning first and second functions:
♦ Let first function be: f(x) = $\small{\sqrt{x^2 - a^2}}$
♦ Let second function be: g(x) = $\small{1}$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~x}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sqrt{x^2 - a^2} \, \left(x\right) \big]}$
• This is the first term.
4. $\small{f'(x)~=~\frac{x}{\sqrt{x^2-a^2}}}$
(The reader must write all steps for this differentiation)
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[\frac{x}{\sqrt{x^2-a^2}}\,\left(x \right) \big]dx}~=~\int{\big[\frac{x^2}{\sqrt{x^2-a^2}} \big]dx}}$
• This can be rearranged as shown below:
$\small{\int{\big[\frac{x^2}{\sqrt{x^2-a^2}} \big]dx}~=~\int{\big[\frac{x^2 - a^2 + a^2}{\sqrt{x^2-a^2}} \big]dx}~=~\int{\big[\frac{x^2 - a^2}{\sqrt{x^2-a^2}}~+~\frac{ a^2}{\sqrt{x^2-a^2}} \big]dx}}$
$\small{~=~\int{\big[\sqrt{x^2-a^2}~+~\frac{ a^2}{\sqrt{x^2-a^2}} \big]dx}}$
$\small{~=~\int{\big[\sqrt{x^2-a^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}} \big]dx}}$
• This is the second term
6. So we get:
$\small{\int{\left[\sqrt{x^2 - a^2} \right]dx}~=~\text{First term - Second term}}$
$\small{~=~\big[x \sqrt{x^2 - a^2} \big]~-~\bigg[\int{\big[\sqrt{x^2-a^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}} \big]dx} \bigg]}$
$\small{~=~x \sqrt{x^2 - a^2} ~-~\int{\big[\sqrt{x^2-a^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}} \big]dx} }$
$\small{\Rightarrow 2\int{\big[\sqrt{x^2-a^2} \big]dx}~=~x \sqrt{x^2 - a^2} ~-~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}} \big]dx} }$
$\small{\Rightarrow \int{\big[\sqrt{x^2-a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 - a^2} ~-~\frac{a^2}{2} \int{\big[\frac{1}{\sqrt{x^2-a^2}} \big]dx} }$
• The last term in the R.H.S can be calculated using formula IV that we saw in section 23.8. Thus we get:
$\small{\int{\big[\sqrt{x^2-a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 - a^2} ~-~\frac{a^2}{2} \log \left|x+\sqrt{x^2 - a^2} \right|~+~\rm{C}}$
II. The integral $\bf{\int{\left[\sqrt{x^2 + a^2} \right]dx}}$
This can be calculated as follows:
1. Assigning first and second functions:
♦ Let first function be: f(x) = $\small{\sqrt{x^2 + a^2}}$
♦ Let second function be: g(x) = $\small{1}$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~x}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sqrt{x^2 + a^2} \, \left(x\right) \big]}$
• This is the first term.
4. $\small{f'(x)~=~\frac{x}{\sqrt{x^2+a^2}}}$
(The reader must write all steps for this differentiation)
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[\frac{x}{\sqrt{x^2+a^2}}\,\left(x \right) \big]dx}~=~\int{\big[\frac{x^2}{\sqrt{x^2+a^2}} \big]dx}}$
• This can be rearranged as shown below:
$\small{\int{\big[\frac{x^2}{\sqrt{x^2+a^2}} \big]dx}~=~\int{\big[\frac{x^2 + a^2 - a^2}{\sqrt{x^2+a^2}} \big]dx}~=~\int{\big[\frac{x^2 + a^2}{\sqrt{x^2-a^2}}~-~\frac{ a^2}{\sqrt{x^2-a^2}} \big]dx}}$
$\small{~=~\int{\big[\sqrt{x^2+a^2}~-~\frac{ a^2}{\sqrt{x^2+a^2}} \big]dx}}$
$\small{~=~\int{\big[\sqrt{x^2+a^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}} \big]dx}}$
• This is the second term
6. So we get:
$\small{\int{\left[\sqrt{x^2 + a^2} \right]dx}~=~\text{First term - Second term}}$
$\small{~=~\big[x \sqrt{x^2 + a^2} \big]~-~\bigg[\int{\big[\sqrt{x^2+a^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}} \big]dx} \bigg]}$
$\small{~=~x \sqrt{x^2 + a^2} ~-~\int{\big[\sqrt{x^2+a^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}} \big]dx} }$
$\small{\Rightarrow 2\int{\big[\sqrt{x^2+a^2} \big]dx}~=~x \sqrt{x^2 + a^2} ~+~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}} \big]dx} }$
$\small{\Rightarrow \int{\big[\sqrt{x^2+a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \int{\big[\frac{1}{\sqrt{x^2+a^2}} \big]dx} }$
• The last term in the R.H.S can be calculated using formula VI that we saw in section 23.8. Thus we get:
$\small{ \int{\big[\sqrt{x^2+a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \log \left|x+\sqrt{x^2 + a^2} \right|~+~\rm{C}}$
III. The integral $\bf{\int{\left[\sqrt{a^2 - x^2} \right]dx}}$
This can be calculated as follows:
1. Assigning first and second functions:
♦ Let first function be: f(x) = $\small{\sqrt{a^2 - x^2}}$
♦ Let second function be: g(x) = $\small{1}$
2. Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~x}$
3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sqrt{a^2 - x^2} \, \left(x\right) \big]}$
• This is the first term.
4. $\small{f'(x)~=~\frac{-x}{\sqrt{a^2-x^2}}}$
(The reader must write all steps for this differentiation)
5. $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[\frac{-x}{\sqrt{a^2-x^2}}\,\left(x \right) \big]dx}~=~\int{\big[\frac{-x^2}{\sqrt{a^2-x^2}} \big]dx}}$
• This can be rearranged as shown below:
$\small{\int{\big[\frac{-x^2}{\sqrt{a^2-x^2}} \big]dx}~=~\int{\big[\frac{a^2 - x^2 - a^2}{\sqrt{a^2-x^2}} \big]dx}~=~\int{\big[\frac{a^2 - x^2}{\sqrt{a^2-x^2}}~-~\frac{ a^2}{\sqrt{a^2-x^2}} \big]dx}}$
$\small{~=~\int{\big[\sqrt{a^2-x^2}~-~\frac{ a^2}{\sqrt{a^2-x^2}} \big]dx}}$
$\small{~=~\int{\big[\sqrt{a^2-x^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{a^2-x^2}} \big]dx}}$
• This is the second term
6. So we get:
$\small{\int{\left[\sqrt{a^2 - x^2} \right]dx}~=~\text{First term - Second term}}$
$\small{~=~\big[x\sqrt{a^2 - x^2} \big]~-~\bigg[\int{\big[\sqrt{a^2-x^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{a^2-x^2}} \big]dx} \bigg]}$
$\small{~=~x\sqrt{a^2 - x^2} ~-~\int{\big[\sqrt{a^2-x^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{a^2-x^2}} \big]dx} }$
$\small{\Rightarrow 2\int{\big[\sqrt{a^2-x^2} \big]dx}~=~x\sqrt{a^2 - x^2} ~+~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}} \big]dx} }$
$\small{\Rightarrow \int{\big[\sqrt{a^2-x^2} \big]dx}~=~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \int{\big[\frac{1}{\sqrt{a^2-x^2}} \big]dx} }$
• The last term in the R.H.S can be calculated using formula V that we saw in section 23.8. Thus we get:
$\small{ \int{\big[\sqrt{a^2-x^2} \big]dx}~=~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}~+~\rm{C}}$
Now we will see some solved examples.
Solved example 23.61
Find $\small{\int{\left[\sqrt{x^2 +2x+ 5} \right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:
$\small{\int{\left[\sqrt{x^2 +2x+ 1+ 4} \right]dx}~=~\int{\left[\sqrt{(x+1)^2+ 4} \right]dx}~=~\int{\left[\sqrt{(x+1)^2+ 2^2} \right]dx}}$
2. Put u = x+1 Then du/dx = 1 ⇒ du = dx
So the given integral becomes: $\small{\int{\left[\sqrt{u^2+ 4} \right]du}}$
3. Now we can use the standard integral:
$\bf{\int{\left[\sqrt{x^2 + a^2} \right]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \log \left|x+\sqrt{x^2 + a^2} \right|~+~\rm{C}}$
Here x = u and a = 2
• So we get:
$\small{\int{\left[\sqrt{u^2 + 2^2} \right]du}~=~\frac{u}{2}\sqrt{u^2 + 2^2} ~+~\frac{2^2}{2} \log \left|u+\sqrt{u^2 + 2^2} \right|~+~\rm{C}}$
4. Substituting for u, we get:
$\small{\int{\left[\sqrt{(x+1)^2 + 2^2} \right]dx}~=~\frac{x+1}{2}\sqrt{(x+1)^2 + 2^2} ~+~\frac{2^2}{2} \log \left|(x+1)+\sqrt{(x+1)^2 + 2^2} \right|~+~\rm{C}}$
$\small{\Rightarrow \int{\left[\sqrt{x^2 + 2x + 5} \right]dx}~=~\frac{x+1}{2}\sqrt{x^2 + 2x + 5} ~+~2 \log \left|(x+1)+\sqrt{x^2 + 2x + 5} \right|~+~\rm{C}}$
Solved Example 23.62
Find $\small{\int{\left[\sqrt{x^2 +4x+ 6} \right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:
$\small{\int{\left[\sqrt{x^2 +4x+ 4+ 2} \right]dx}~=~\int{\left[\sqrt{(x+2)^2+ 2} \right]dx}~=~\int{\left[\sqrt{(x+1)^2+ (\sqrt{2})^2} \right]dx}}$
2. Put u = x+2 Then du/dx = 1 ⇒ du = dx
So the given integral becomes: $\small{\int{\left[\sqrt{u^2+ (\sqrt{2})^2} \right]du}}$
3. Now we can use the standard integral:
$\bf{\int{\left[\sqrt{x^2 + a^2} \right]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \log \left|x+\sqrt{x^2 + a^2} \right|~+~\rm{C}}$
Here x = u and a = √2
• So we get:
$\small{\int{\left[\sqrt{u^2 + (\sqrt{2})^2} \right]du}~=~\frac{u}{2}\sqrt{u^2 + (\sqrt{2})^2} ~+~\frac{(\sqrt{2})^2}{2} \log \left|u+\sqrt{u^2 + (\sqrt{2})^2} \right|~+~\rm{C}}$
4. Substituting for u, we get:
$\small{\int{\left[\sqrt{(x+2)^2 + (\sqrt{2})^2} \right]dx}~=~\frac{x+2}{2}\sqrt{(x+1)^2 + (\sqrt{2})^2} ~+~\frac{(\sqrt{2})^2}{2} \log \left|(x+2)+\sqrt{(x+2)^2 + (\sqrt{2})^2} \right|~+~\rm{C}}$
$\small{\Rightarrow \int{\left[\sqrt{x^2 + 4x + 6} \right]dx}~=~\frac{x+2}{2}\sqrt{x^2 + 4x + 6} ~+~ \log \left|x+2+\sqrt{x^2 + 4x + 6} \right|~+~\rm{C}}$
We
have seen the three standard integrals.
In the next section, we will see a few more solved examples.
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