In the previous section, we completed a discussion on the application of trigonometric identities in the process of integration. We saw some solved examples also. In this section, we will see the integrals of some particular functions.
First we will see six basic results. They can be used as formulas for solving complex problems.
Formula I
$\small{\int{\left[\frac{dx}{x^2\,-\,m^2} \right]}\,=\,\frac{1}{2m}\log \left | \frac{x-m}{x+m} \right |\,+\,C}$
Proof can be written in 3 steps:
1. We can rearrange the L.H.S as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{1}{x^2\,-\,m^2}} & {~=~} &{\frac{1}{(x+m)(x-m)}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\frac{(x+m)-(x-m)}{2m(x+m)(x-m)}} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{(x+m)}{2m(x+m)(x-m)}~-~\frac{(x-m)}{2m(x+m)(x-m)}} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{1}{2m(x-m)}~-~\frac{1}{2m(x+m)}} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\frac{1}{2m}\,\left[\frac{1}{x-m}~-~\frac{1}{x+m} \right]} \\
\end{array}}$
2. So we want:
$\small{\int{\left[\frac{dx}{x^2\,-\,m^2} \right]dx}\,=\,\int{\left[\frac{1}{2m}\,\left[\frac{1}{x-m}~-~\frac{1}{x+m} \right] \right]dx}}$
= $\small{\int{\left[\frac{1}{2m}\,\left[\frac{1}{x-m} \right] \right]dx}\,-\,\int{\left[\frac{1}{2m}\,\left[\frac{1}{x+m} \right] \right]dx}}$
3. This integration gives:
$\small{\frac{1}{2m} \log|x-m|~-~\frac{1}{2m} \log|x+m|\,+\,\rm{C}~=~\frac{1}{2m} \log \left|\frac{x-m}{x+m} \right|\,+\,\rm{C}}$
Formula II
$\small{\int{\left[\frac{dx}{m^2\,-\,x^2} \right]}\,=\,\frac{1}{2m}\log \left | \frac{m+x}{m-x} \right |\,+\,C}$
Proof can be written in 3 steps:
1. We can rearrange the L.H.S as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{1}{m^2\,-\,x^2}} & {~=~} &{\frac{1}{(m+x)(m-x)}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\frac{(m+x)+(m-x)}{2m(m+x)(m-x)}} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{(m+x)}{2m(m+x)(m-x)}~+~\frac{(m-x)}{2m(m+x)(m-x)}} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{1}{2m(m-x)}~+~\frac{1}{2m(m+x)}} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\frac{1}{2m}\,\left[\frac{1}{m-x}~+~\frac{1}{m+x} \right]} \\
\end{array}}$
2. So we want:
$\small{\int{\left[\frac{dx}{m^2\,-\,x^2} \right]dx}\,=\,\int{\left[\frac{1}{2m}\,\left[\frac{1}{m-x}~+~\frac{1}{m+x} \right] \right]dx}}$
= $\small{\int{\left[\frac{1}{2m}\,\left[\frac{1}{m-x} \right] \right]dx}\,+\,\int{\left[\frac{1}{2m}\,\left[\frac{1}{m+x} \right] \right]dx}}$
3. This integration gives:
$\small{\frac{-1}{2m} \log|m-x|~+~\frac{1}{2m} \log|m+x|\,+\,\rm{C}~=~\frac{1}{2m} \log \left|\frac{m+x}{m-x} \right|\,+\,\rm{C}}$
Formula III
$\small{\int{\left[\frac{dx}{x^2\,+\,m^2} \right]}\,=\,\frac{1}{m} \tan^{-1}\frac{x}{m}\,+\,\rm{C}}$
Proof can be written in 3 steps:
1. Let x = m tan 𝜃.
Then we get: $\small{\frac{dx}{d \theta}\,=\,m \sec^2 \theta}$
⇒ $\small{dx\,=\,\,m \sec^2 \theta\,d \theta}$
2. So we want:
$\small{\int{\left[\frac{dx}{x^2\,+\,m^2} \right]dx}\,=\,\int{\left[\frac{m \sec^2 \theta\,d \theta}{(m \tan \theta)^2\,+\,m^2} \right]}}$
$\small{~=~\int{\left[\frac{m \sec^2 \theta\,d \theta}{m^2 \tan^2 \theta\,+\,m^2} \right]}~=~\int{\left[\frac{m \sec^2 \theta\,d \theta}{m^2 (\tan^2 \theta\,+\,1)} \right]}}$
$\small{~=~\int{\left[\frac{m \sec^2 \theta\,d \theta}{m^2 (\sec^2 \theta)} \right]}~=~\int{\left[\frac{d \theta}{m} \right]}}$
3. This integration gives:
$\small{\frac{\theta}{m}\,+\,\rm{C}~=~\frac{1}{m} \tan^{-1}\frac{x}{m}\,+\,\rm{C}}$
Formula IV
$\small{\int{\left[\frac{dx}{\sqrt{x^2\,-\,m^2}} \right]}\,=\, \log \left|x\,+\,\sqrt{x^2\,-\,m^2} \right| \,+\,\rm{C}}$
Proof can be written in 4 steps:
1. Let x = m sec 𝜃.
Then we get: $\small{\frac{dx}{d \theta}\,=\,m \sec \theta \tan \theta}$
⇒ $\small{dx\,=\,\,m \sec \theta \tan \theta \,d \theta}$
2. So we want:
$\small{\int{\left[\frac{dx}{\sqrt{x^2\,-\,m^2}} \right]dx}\,=\,\int{\left[\frac{m \sec \theta \tan \theta\,d \theta}{\sqrt{(m \sec \theta)^2\,-\,m^2}} \right]}}$
$\small{~=~\int{\left[\frac{m \sec \theta \tan \theta\,d \theta}{\sqrt{m^2 \sec^2 \theta\,-\,m^2}} \right]}~=~\int{\left[\frac{m \sec \theta \tan \theta\,d \theta}{\sqrt{m^2 (\sec^2 \theta\,-\,1)}} \right]}}$
$\small{~=~\int{\left[\frac{m \sec \theta \tan \theta\,d \theta}{\sqrt{m^2 \tan^2 \theta}} \right]}~=~\int{\left[\frac{m \sec \theta \tan \theta\,d \theta}{m \tan \theta} \right]}}$
$\small{~=~\int{\left[\sec \theta \right]d \theta}}$
3. This integration gives:
$\small{\log |\sec \theta\,+\,\tan \theta|\,+\,\rm{C_1}}$
4. From the above result, we can eliminate 𝜃 as shown below:
• In (1), we wrote: x = m sec 𝜃.
So sec 𝜃 = x/m.
Then $\small{\tan \theta \,=\, \sqrt{\sec^2 \theta \,-\,1} \,=\,\sqrt{\frac{x^2}{m^2}\,-\,1}}$
• Now we get:
$\small{\log |\sec \theta\,+\,\tan \theta|\,+\,\rm{C_1}}$
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{{}} & {~=~} &{\log \left |\frac{x}{m}\,+\, \sqrt{\frac{x^2}{m^2}\,-\,1} \right |\,+\,\rm{C_1}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\log \left |\frac{x}{m}\,+\, \sqrt{\frac{x^2\,-\,m^2}{m^2}} \right |\,+\,\rm{C_1}} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\log \left |\frac{x\,+\,\sqrt{x^2 - m^2}}{m} \right |\,+\,\rm{C_1}} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\log \left |x\,+\,\sqrt{x^2 - m^2}\right |\,-\,\log \left |m \right |\,+\,\rm{C_1}} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\log \left |x\,+\,\sqrt{x^2 - m^2}\right |\,+\,\rm{C}} \\
\end{array}}$
◼ Remarks:
5 (magenta color): log|m| is a constant. So it can be combined with C1 to form a new constant C.
Formula V
$\small{\int{\left[\frac{dx}{\sqrt{m^2\,-\,x^2}} \right]}\,=\, \sin^{-1} \frac{x}{m} \,+\,\rm{C}}$
Proof can be written in 3 steps:
1. Let x = m sin 𝜃.
Then we get: $\small{\frac{dx}{d \theta}\,=\,m \cos \theta }$
⇒ $\small{dx\,=\,\,m \cos \theta \,d \theta}$
2. So we want:
$\small{\int{\left[\frac{dx}{\sqrt{m^2\,-\,x^2}} \right]dx}\,=\,\int{\left[\frac{m \cos \theta\,d \theta}{\sqrt{m^2\,-\,(m \sin \theta)^2}} \right]}}$
$\small{~=~\int{\left[\frac{m \cos \theta\,d \theta}{\sqrt{m^2 \,-\,m^2 \sin^2 \theta}} \right]}~=~\int{\left[\frac{m \cos \theta\,d \theta}{\sqrt{m^2(1 \,-\,\sin^2 \theta)}} \right]}}$
$\small{~=~\int{\left[\frac{m \cos \theta\,d \theta}{\sqrt{m^2 \cos^2 \theta}} \right]}~=~\int{\left[1 \right]d \theta}}$
3. This integration gives:
$\small{\theta \,+\,\rm{C}\,=\,\sin^{-1}\frac{x}{m}\,+\,\rm{C}}$
Formula VI
$\small{\int{\left[\frac{dx}{\sqrt{x^2\,+\,m^2}} \right]}\,=\, \log \left|x\,+\,\sqrt{x^2\,+\,m^2} \right| \,+\,\rm{C}}$
Proof can be written in 4 steps:
1. Let x = m tan 𝜃.
Then we get: $\small{\frac{dx}{d \theta}\,=\,m \sec^2 \theta}$
⇒ $\small{dx\,=\,\,m \sec^2 \theta \,d \theta}$
2. So we want:
$\small{\int{\left[\frac{dx}{\sqrt{x^2\,+\,m^2}} \right]dx}\,=\,\int{\left[\frac{m \sec^2 \theta \,d \theta}{\sqrt{(m \tan \theta)^2\,+\,m^2}} \right]}}$
$\small{~=~\int{\left[\frac{m \sec^2 \theta \,d \theta}{\sqrt{m^2 \tan^2 \theta\,+\,m^2}} \right]}~=~\int{\left[\frac{m \sec^2 \theta\,d \theta}{\sqrt{m^2 (\tan^2 \theta\,+\,1)}} \right]}}$
$\small{~=~\int{\left[\frac{m \sec^2 \theta\,d \theta}{\sqrt{m^2 (\sec^2 \theta)}} \right]}~=~\int{\left[\sec \theta \right] d \theta}}$
3. This integration gives:
$\small{\log |\sec \theta\,+\,\tan \theta|\,+\,\rm{C_1}}$
4. From the above result, we can eliminate 𝜃 as shown below:
• In (1), we wrote: x = m tan 𝜃.
So tan 𝜃 = x/m.
Then $\small{\sec \theta \,=\, \sqrt{\tan^2 \theta \,+\,1} \,=\,\sqrt{\frac{x^2}{m^2}\,+\,1}}$
• Now we get:
$\small{\log |\sec \theta\,+\,\tan \theta|\,+\,\rm{C_1}}$
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{{}} & {~=~} &{\log \left |\sqrt{\frac{x^2}{m^2}\,+\,1}\,+\,\frac{x}{m} \right |\,+\,\rm{C_1}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\log \left |\sqrt{\frac{x^2\,+\,m^2}{m^2}}\,+\, \frac{x}{m} \right |\,+\,\rm{C_1}} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\log \left |\frac{\sqrt{x^2 + m^2}\,+\,x}{m} \right |\,+\,\rm{C_1}} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\log \left |x\,+\,\sqrt{x^2 + m^2}\right |\,-\,\log \left |m \right |\,+\,\rm{C_1}} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\log \left |x\,+\,\sqrt{x^2 + m^2}\right |\,+\,\rm{C}} \\
\end{array}}$
◼ Remarks:
5 (magenta color): log|m| is a constant. So it can be combined with C1 to form a new constant C.
We have seen six basic integrals. Using those, we can derive four more integrals. We will see them in the next section.
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