In the previous section, we saw the derivation of three standard integrals. We saw some solved examples also. In this section, we will see a few more solved examples.
Solved Example 23.63
Find $\small{\int{\left[\sqrt{4 - x^2} \right]dx}}$
Solution:
1. The given integral can be written as: $\small{\int{\left[\sqrt{2^2 - x^2} \right]dx}}$
2. Now we can use the standard integral:
$\bf{\int{\left[\sqrt{a^2 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}~+~\rm{C}}$
Here a = 2
• So we get:
$\small{\int{\left[\sqrt{2^2 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{2^2 - x^2} ~+~\frac{2^2}{2} \sin^{-1}\frac{x}{2}~+~\rm{C}}$
$\small{\Rightarrow \int{\left[\sqrt{4 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{4 - x^2} ~+~2 \sin^{-1}\frac{x}{2}~+~\rm{C}}$
Solved Example 23.64
Find $\small{\int{\left[\sqrt{1 - 4 x^2} \right]dx}}$
Solution:
1. The given integral can be written as:
$\small{\int{\left[\sqrt{1^2 - (2x)^2} \right]dx}}$
2. Put u = 2x Then du/dx = 2 ⇒ du = 2 dx
• So the given integral becomes:
$\small{\int{\left[\sqrt{1^2 - (2x)^2} \right]dx}~=~\int{\left[\frac{2 \sqrt{1^2 - (2x)^2}}{2} \right]dx}}$
$\small{~=~\int{\left[\frac{ \sqrt{1^2 - u^2}}{2} \right]du}~=~\frac{1}{2} \int{\left[\sqrt{1^2 - u^2} \right]dx}}$
3. Now we can use the standard integral:
$\bf{\int{\left[\sqrt{a^2 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}~+~\rm{C}}$
Here a = 1
• So we get:
$\small{\frac{1}{2}\int{\left[\sqrt{1^2 - u^2} \right]du}~=~~\frac{u}{4}\sqrt{1^2 - u^2} ~+~\frac{1^2}{4} \sin^{-1}\frac{u}{1}~+~\rm{C}}$
4. Substituting for u, we get:
$\small{\frac{1}{2}\int{\left[\sqrt{1^2 - (2x)^2} \right]dx}~=~~\frac{2x}{4}\sqrt{1^2 - (2x)^2} ~+~\frac{1}{4} \sin^{-1}(2x)~+~\rm{C}}$
$\small{\Rightarrow \frac{1}{2}\int{\left[\sqrt{1 - 4x^2} \right]dx}~=~~\frac{x}{2}\sqrt{1 - 4x^2} ~+~\frac{1}{4} \sin^{-1}(2x)~+~\rm{C}}$
Solved Example 23.65
Find $\small{\int{\left[\sqrt{x^2 + 4x + 1} \right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:
$\small{\int{\left[\sqrt{x^2 +4x+ 4 - 3} \right]dx}~=~\int{\left[\sqrt{(x+2)^2 - (\sqrt{3})^2} \right]dx}}$
2. Put u = x+2 Then du/dx = 1 ⇒ du = dx
So the given integral becomes: $\small{\int{\left[\sqrt{u^2 - (\sqrt{3})^2} \right]du}}$
3. Now we can use the standard integral:
$\bf{\int{\big[\sqrt{x^2-a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 - a^2} ~-~\frac{a^2}{2} \log \left|x+\sqrt{x^2 - a^2} \right|~+~\rm{C}}$
Here x = u and a = √3
• So we get:
$\small{\int{\left[\sqrt{u^2 - (\sqrt{3})^2} \right]du}~=~\frac{u}{2}\sqrt{u^2 - (\sqrt{3})^2} ~-~\frac{(\sqrt{3})^2}{2} \log \left|u+\sqrt{u^2 - (\sqrt{3})^2} \right|~+~\rm{C}}$
4. Substituting for u, we get:
$\small{\int{\left[\sqrt{(x+2)^2 - (\sqrt{3})^2} \right]dx}~=~\frac{x+2}{2}\sqrt{(x+2)^2 - (\sqrt{3})^2} ~-~\frac{(\sqrt{3})^2}{2} \log \left|(x+2)-\sqrt{(x+2)^2 + (\sqrt{3})^2} \right|~+~\rm{C}}$
$\small{\Rightarrow \int{\left[\sqrt{x^2 + 4x + 1} \right]dx}~=~\frac{x+2}{2}\sqrt{x^2 + 4x + 1} ~-~ \frac{3}{2} \log \left|x+2-\sqrt{x^2 + 4x + 1} \right|~+~\rm{C}}$
Solved Example 23.66
Find $\small{\int{\left[\sqrt{x^2 + 4x - 5} \right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:
$\small{\int{\left[\sqrt{x^2 +4x+ 4 - 9} \right]dx}~=~\int{\left[\sqrt{(x+2)^2 - 3^2} \right]dx}}$
2. Put u = x+2 Then du/dx = 1 ⇒ du = dx
So the given integral becomes: $\small{\int{\left[\sqrt{u^2 - 3^2} \right]du}}$
3. Now we can use the standard integral:
$\bf{\int{\big[\sqrt{x^2-a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 - a^2} ~-~\frac{a^2}{2} \log \left|x+\sqrt{x^2 - a^2} \right|~+~\rm{C}}$
Here x = u and a = 3
• So we get:
$\small{\int{\left[\sqrt{u^2 - 3^2} \right]du}~=~\frac{u}{2}\sqrt{u^2 - 3^2} ~-~\frac{3^2}{2} \log \left|u+\sqrt{u^2 - 3^2} \right|~+~\rm{C}}$
4. Substituting for u, we get:
$\small{\int{\left[\sqrt{(x+2)^2 - 3^2} \right]dx}}$
$\small{~=~\frac{x+2}{2}\sqrt{(x+2)^2 - 3^2} ~-~\frac{3^2}{2} \log \left|(x+2)-\sqrt{(x+2)^2 + 3^2} \right|~+~\rm{C}}$
$\small{\Rightarrow \int{\left[\sqrt{x^2 + 4x - 5} \right]dx}}$
$\small{~=~\frac{x+2}{2}\sqrt{x^2 + 4x - 5} ~-~ \frac{9}{2} \log \left|x+2-\sqrt{x^2 + 4x - 5} \right|~+~\rm{C}}$
Solved Example 23.67
Find $\small{\int{\left[\sqrt{3 - 2x - x^2}\right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:
$\small{\int{\left[\sqrt{-(x^2 +2x- 3)} \right]dx}~=~\int{\left[\sqrt{-[x^2 +2x+1-4]}\right]dx}}$
$\small{~=~\int{\left[\sqrt{-\left[(x+1)^2-2^2\right]}\right]dx}~=~\int{\left[\sqrt{2^2 - (x+1)^2}\right]dx}}$
2. Put u = x+1 Then du/dx = 1 ⇒ du = dx
So the given integral becomes: $\small{\int{\left[\sqrt{2^2 - u^2} \right]du}}$
3. Now we can use the standard integral:
$\bf{\int{\left[\sqrt{a^2 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}~+~\rm{C}}$
Here a = 2 and x = u
• So we get:
$\small{\int{\left[\sqrt{2^2 - u^2} \right]dx}~=~~\frac{u}{2}\sqrt{2^2 - u^2} ~+~\frac{2^2}{2} \sin^{-1}\frac{u}{2}~+~\rm{C}}$
4. Substituting for u, we get:
$\small{\int{\left[\sqrt{2^2 - (x+1)^2} \right]dx}}$
$\small{~=~~\frac{x+1}{2}\sqrt{2^2 - (x+1)^2} ~+~\frac{2^2}{2} \sin^{-1}\frac{x+1}{2}~+~\rm{C}}$
$\small{\Rightarrow \int{\left[\sqrt{3-2x-x^2} \right]dx}}$
$\small{~=~\frac{x+1}{2}\sqrt{3-2x-x^2}~+~2 \sin^{-1}\frac{x+1}{2}~+~\rm{C}}$
Solved Example 23.68
Find $\small{\int{\left[\sqrt{1 - 4x - x^2}\right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:
$\small{\int{\left[\sqrt{-(x^2 +4x- 1)} \right]dx}~=~\int{\left[\sqrt{-[x^2 +4x+4-5]}\right]dx}}$
$\small{~=~\int{\left[\sqrt{-\left[(x+2)^2-(\sqrt{5})^2\right]}\right]dx}~=~\int{\left[\sqrt{(\sqrt{5})^2 - (x+2)^2}\right]dx}}$
2. Put u = x+2 Then du/dx = 1 ⇒ du = dx
So the given integral becomes: $\small{\int{\left[\sqrt{(\sqrt{5})^2 - u^2} \right]du}}$
3. Now we can use the standard integral:
$\bf{\int{\left[\sqrt{a^2 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}~+~\rm{C}}$
Here a = √5 and x = u
• So we get:
$\small{\int{\left[\sqrt{(\sqrt{5})^2 - u^2} \right]dx}~=~~\frac{u}{2}\sqrt{(\sqrt{5})^2 - u^2} ~+~\frac{(\sqrt{5})^2}{2} \sin^{-1}\frac{u}{\sqrt{5}}~+~\rm{C}}$
4. Substituting for u, we get:
$\small{\int{\left[\sqrt{(\sqrt{5})^2 - (x+2)^2} \right]dx}}$
$\small{~=~~\frac{x+2}{2}\sqrt{(\sqrt{5})^2 - (x+2)^2} ~+~\frac{(\sqrt{5})^2}{2} \sin^{-1}\frac{x+2}{\sqrt{5}}~+~\rm{C}}$
$\small{\Rightarrow \int{\left[\sqrt{1-4x-x^2} \right]dx}}$
$\small{~=~\frac{x+2}{2}\sqrt{1-4x-x^2}~+~\frac{5}{2} \sin^{-1}\frac{x+2}{\sqrt{5}}~+~\rm{C}}$
The link below gives a few more solved examples:
Exercise 23.7
We
have completed a discussion on the three standard integrals.
In the next section, we will see definite integrals.
Previous
Contents
Copyright©2025 Higher secondary mathematics.blogspot.com
No comments:
Post a Comment