Tuesday, June 24, 2025

23.27 - Solved Examples on Properties of Definite Integrals

In the previous section, we saw some properties of definite integrals. We saw some solved examples also. In this section, we will see a few more solved examples.

First we will write the method to test whether a given function is odd, even, neither:

Even: $\small{f(-x) \,=\,f(x)}$ 

Odd: $\small{f(-x) \,=\,-f(x)}$ 

Neither: $\small{f(-x) \, \ne \,f(x)}$  and

              $\small{f(-x) \, \ne \,-f(x)}$

Let us see an example:

Check whether the function $\small{f(x)\,=\,x^2 + x}$ is even, odd or neither.

Solution:
1. $\small{f(-x)\,=\,(-x)^2 + (-x)\,=\,x^2 \,-\,x}$ 

2. $\small{-f(x)\,=\,-x^2\,-\,x}$

3. We see that:

$\small{f(-x) \, \ne \,f(x)}$ and  

$\small{f(-x) \, \ne \,-f(x)}$

So the given function is neither even nor odd


Now we will proceed with the solved examples:

Solved Example 23.104
Evaluate $\small{\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}{\left[\sin^2 x \right]dx}}$
Solution:
1. The limits are $\small{\frac{-\pi}{4}~\rm{and}~\frac{\pi}{4}}$.
So we check whether $\small{f(x)=\sin^2 x}$ is even or odd.

(a) $\small{f(-x)\,=\,\sin^2(-x)\,=\,\left(\sin(-x) \right)^2}$ 

$\small{\,=\,\left(-\sin(x) \right)^2\,=\,\sin^2 x}$ 

$\small{\because \sin(-x) = -\sin x }$

(b) $\small{-f(x)\,=\,-\sin^2 x}$

(c) We see that:

$\small{f(-x) \,=\,f(x)}$

• So the given f(x) is an even function.

2. Since the given function is even, we can apply:

$\bf{{\rm{P_7~(i)}}:}$

$\bf{\int_{-a}^{a}{\left[f(x) \right]dx}~=~2 \int_0^a{\left[f(x) \right]dx}}$
   ♦ If $\small{f}$ is an even function

• We can write:

$\small{\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}{\left[\sin^2 x \right]dx}\,=\,2 \int_{0}^{\frac{\pi}{4}}{\left[\sin^2 x \right]dx}}$

3. Let $\small{I~=~\,2 \int_{0}^{\frac{\pi}{4}}{\left[\sin^2 x \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~2 \int{\left[\sin^2 x \right]dx}~=~2 \int{\left[\frac{1 - \cos(2x)}{2}\right]dx}}$

$\small{~=~2 \int{\left[\frac{1}{2}\right]dx}~+~2 \int{\left[\frac{- \cos(2x)}{2}\right]dx}}$

$\small{~=~2 \left[\frac{x}{2}\right]~+~2 \left[\frac{- \sin(2x)}{2(2)}\right]}$

$\small{~=~x~-~ \frac{\sin(2x)}{2}}$

4. Therefore,

$\small{I~=~F(\frac{\pi}{4})\,-\,F(0)~=~\left[x~-~ \frac{\sin(2x)}{2} \right]_{0}^{\frac{\pi}{4}}}$

$\small{~=~\left[\frac{\pi}{4}~-~ \frac{\sin(\frac{\pi}{2})}{2} \right]~-~\left[0~-~ \frac{\sin(0)}{2} \right]}$

$\small{~=~\left[\frac{\pi}{4}~-~ \frac{1}{2} \right]~-~\left[0~-~ \frac{0}{2} \right]~=~\frac{\pi}{4}~-~ \frac{1}{2}}$

Solved Example 23.105
Evaluate $\small{\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}{\left[\sin^2 x \right]dx}}$
Solution:
1. The limits are $\small{\frac{-\pi}{2}~\rm{and}~\frac{\pi}{2}}$.
So we check whether $\small{f(x)=\sin^2 x}$ is even or odd.

(a) $\small{f(-x)\,=\,\sin^2(-x)\,=\,\left(\sin(-x) \right)^2}$

$\small{\,=\,\left(-\sin(x) \right)^2\,=\,\sin^2 x}$

$\small{\because \sin(-x) = -\sin x }$

(b) $\small{-f(x)\,=\,-\sin^2 x}$

(c) We see that:

$\small{f(-x) \,=\,f(x)}$

• So the given f(x) is an even function.

2. Since the given function is even, we can apply:

$\bf{{\rm{P_7~(i)}}:}$

$\bf{\int_{-a}^{a}{\left[f(x) \right]dx}~=~2 \int_0^a{\left[f(x) \right]dx}}$
   ♦ If $\small{f}$ is an even function

• We can write:

$\small{\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}{\left[\sin^2 x \right]dx}\,=\,2 \int_{0}^{\frac{\pi}{2}}{\left[\sin^2 x \right]dx}}$

3. Let $\small{I~=~\,2 \int_{0}^{\frac{\pi}{2}}{\left[\sin^2 x \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~2 \int{\left[\sin^2 x \right]dx}~=~2 \int{\left[\frac{1 - \cos(2x)}{2}\right]dx}}$

$\small{~=~2 \int{\left[\frac{1}{2}\right]dx}~+~2 \int{\left[\frac{- \cos(2x)}{2}\right]dx}}$

$\small{~=~2 \left[\frac{x}{2}\right]~+~2 \left[\frac{- \sin(2x)}{2(2)}\right]}$

$\small{~=~x~-~ \frac{\sin(2x)}{2}}$

4. Therefore,

$\small{I~=~F(\frac{\pi}{2})\,-\,F(0)~=~\left[x~-~ \frac{\sin(2x)}{2} \right]_{0}^{\frac{\pi}{2}}}$

$\small{~=~\left[\frac{\pi}{2}~-~ \frac{\sin(\pi)}{2} \right]~-~\left[0~-~ \frac{\sin(0)}{2} \right]}$

$\small{~=~\left[\frac{\pi}{2}~-~ \frac{0}{2} \right]~-~\left[0~-~ \frac{0}{2} \right]~=~\frac{\pi}{2}}$

Solved Example 23.106
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\cos^2 x \right]dx}}$
Solution:
1. The limits are $\small{0~\rm{and}~\frac{\pi}{2}}$.

$\small{\frac{\pi}{2}}$ is $\small{\left(2 \times\frac{\pi}{4} \right)}$

So we can apply:

$\bf{{\rm{P_5}}:\int_0^{2a}{\left[f(x) \right]dx}~=~\int_0^a{\left[f(x) \right]dx}~+~\int_0^a{\left[f(2a-x) \right]dx}}$

2. We get:

$\small{\int_{0}^{\frac{\pi}{2}}{\left[\cos^2 x \right]dx}=\int_{0}^{\frac{\pi}{4}}{\left[\cos^2 x \right]dx}+\int_{0}^{\frac{\pi}{4}}{\left[\cos^2 \left( \frac{\pi}{2}-x \right) \right]dx}}$

$\small{=\int_{0}^{\frac{\pi}{4}}{\left[\cos^2 x \right]dx}+\int_{0}^{\frac{\pi}{4}}{\left[\sin^2 x \right]dx}}$

$\small{=\int_{0}^{\frac{\pi}{4}}{\left[\cos^2 x + \sin^2 x \right]dx}=\int_{0}^{\frac{\pi}{4}}{\left[1 \right]dx}}$

$\small{=\left[x \right]_{0}^{\frac{\pi}{4}}dx~=~\frac{\pi}{4}}$

Solved Example 23.107
Evaluate $\small{\int_{0}^{\pi}{\left[\frac{x \sin(x)}{1 + \cos^2 (x)} \right]dx}}$
Solution:
1. Here we can apply:

$\bf{{\rm{P_4}}:\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[f(a-x) \right]dx}}$

We get: $\small{I~=~\int_{0}^{\pi}{\left[\frac{x \sin(x)}{1 + \cos^2 (x)} \right]dx}}$

$\small{~=~\int_{0}^{\pi}{\left[\frac{(\pi - x) \sin(\pi - x)}{1 + \cos^2 (\pi - x)} \right]dx}=\int_{0}^{\pi}{\left[\frac{(\pi - x) \sin(x)}{1 + \left(-\cos(x) \right)^2} \right]dx}}$

$\small{~=\int_{0}^{\pi}{\left[\frac{(\pi - x) \sin(x)}{1 + \cos^2(x)} \right]dx}=\int_{0}^{\pi}{\left[\frac{\pi \sin(x)}{1 + \cos^2(x)} \right]dx}-\int_{0}^{\pi}{\left[\frac{x \sin(x)}{1 + \cos^2(x)} \right]dx}}$

$\small{~=\int_{0}^{\pi}{\left[\frac{\pi \sin(x)}{1 + \cos^2(x)} \right]dx}-I}$

$\small{\Rightarrow 2I~=~\int_{0}^{\pi}{\left[\frac{\pi \sin(x)}{1 + \cos^2(x)} \right]dx}}$

$\small{\Rightarrow I~=~\frac{\pi}{2} \int_{0}^{\pi}{\left[\frac{\sin(x)}{1 + \cos^2(x)} \right]dx}}$

3. First we find the indefinite integral $\small{F~=~\frac{\pi}{2} \int{\left[\frac{\sin(x)}{1 + \cos^2(x)} \right]dx}}$.

• Put u = cos x

$\small{\Rightarrow \frac{du}{dx}=-\sin x \Rightarrow -\sin x \,dx = du}$

• Now F can be written as:

$\small{F=\frac{\pi}{2} \int{\left[\frac{(-1)(-1)\sin(x)}{1 + \cos^2(x)} \right]dx}=\frac{\pi}{2} \int{\left[\frac{(-1)}{1 + u^2} \right]du}}$

• This integration gives:

$\small{(-1)\left(\frac{\pi}{2} \right)\left(\frac{1}{1} \right) \tan^{-1}\frac{u}{1}~=~(-1)\left(\frac{\pi}{2} \right) \tan^{-1}u}$

4. We wrote: u = cos x

When x approach the lower limit zero, u approach 1

When x approach the upper limit $\small{\pi}$, u approach −1.

5. So the given integral can be written as:

$\small{I~=~\frac{\pi}{2} \int_{0}^{\pi}{\left[\frac{\sin(x)}{1 + \cos^2(x)} \right]dx}=\frac{\pi}{2} \int_1^{-1}{\left[\frac{(-1)}{1 + u^2} \right]du}}$

$\small{~=(-1)\frac{\pi}{2} \int_{-1}^{1}{\left[\frac{(-1)}{1 + u^2} \right]du}~\rm{[by~P_1]}~=~\frac{\pi}{2} \int_{-1}^{1}{\left[\frac{1}{1 + u^2} \right]du}}$

$\small{~=~(2)\frac{\pi}{2} \int_{0}^{1}{\left[\frac{1}{1 + u^2} \right]du}~\rm{[by~P_7,~since ~this~is~even~function]}}$

$\small{~=~\pi \int_{0}^{1}{\left[\frac{1}{1 + u^2} \right]du}}$

6. Therefore,

$\small{I~=~F(1)\,-\,F(0)~=~\left[(\pi)\tan^{-1}u \right]_{0}^{1}}$

$\small{~=~\left[(\pi) \tan^{-1}1 \right]~-~\left[(\pi) \tan^{-1}0 \right]}$

$\small{~=~\left[(\pi) \frac{\pi}{4} \right]~-~\left[(\pi) 0 \right]~=~\frac{\pi^2}{4}}$

Solved Example 23.108
Evaluate $\small{\int_{-1}^{1}{\left[\sin^5 (x) \, \cos^4(x) \right]dx}}$
Solution:
1. The limits are −1 and 1.
So we check whether $\small{f(x)=\sin^5 (x) \, \cos^4(x)}$ is even or odd.

(a) $\small{f(-x)\,=\,\sin^5 (-x) \, \cos^4(-x)\,=\,\left(\sin(-x) \right)^5\,\left(\cos(-x) \right)^4}$

$\small{\,=\,\left(-\sin(x) \right)^5\,\left(\cos(x) \right)^4\,=\,-\sin^5 x \, \cos^4 x}$

$\small{\because \sin(-x) = -\sin x ~~\rm{and}~~\cos(-x) = \cos x}$

(b) $\small{-f(x)\,=\,-\sin^5 (x) \, \cos^4(x)}$

(c) We see that:

$\small{f(-x) \,=\,-f(x)}$

• So the given f(x) is an odd function.

2. Since the given function is odd, we can apply:

$\bf{{\rm{P_7~(ii)}}:}$

$\bf{\int_{-a}^{a}{\left[f(x) \right]dx}~=~0}$
   ♦ If $\small{f}$ is an odd function

• We can write: $\small{\int_{-1}^{1}{\left[\sin^5 (x) \, \cos^4(x) \right]dx}~=~0}$

Solved Example 23.109
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^4 x}{\sin^4 x + \cos^4 x} \right]dx}}$
Solution:
1. The limits are $\small{0~\rm{and}~\frac{\pi}{2}}$. So we will apply:

$\bf{{\rm{P_4}}:\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[f(a-x) \right]dx}}$

We get:
$\small{I~=~\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^4 x}{\sin^4 x + \cos^4 x} \right]dx}=\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^4 \left(\frac{\pi}{2} - x \right)}{\sin^4 \left(\frac{\pi}{2} - x \right) + \cos^4 \left(\frac{\pi}{2} - x \right)} \right]dx}}$

$\small{~=~\int_{0}^{\frac{\pi}{2}}{\left[\frac{\cos^4 x}{\cos^4 x + \sin^4 x} \right]dx}}$

2. Now, I + I = 2I =

$\small{\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^4 x}{\sin^4 x + \cos^4 x} \right]dx}+\int_{0}^{\frac{\pi}{2}}{\left[\frac{\cos^4 x}{\cos^4 x + \sin^4 x} \right]dx}=\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^4 x + \cos^4 x}{\sin^4 x + \cos^4 x} \right]dx}}$

$\small{~=\int_{0}^{\frac{\pi}{2}}{\left[1 \right]dx}=[x]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}}$

Therefore, $\small{I = \frac{\pi}{4}}$

Solved Example 23.110
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\frac{\cos^5 x}{\sin^5 x + \cos^5 x} \right]dx}}$
Solution:
1. The limits are $\small{0~\rm{and}~\frac{\pi}{2}}$. So we will apply:

$\bf{{\rm{P_4}}:\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[f(a-x) \right]dx}}$

We get:
$\small{I~=~\int_{0}^{\frac{\pi}{2}}{\left[\frac{\cos^5 x}{\sin^5 x + \cos^5 x} \right]dx}=\int_{0}^{\frac{\pi}{2}}{\left[\frac{\cos^5 \left(\frac{\pi}{2} - x \right)}{\sin^5 \left(\frac{\pi}{2} - x \right) + \cos^5 \left(\frac{\pi}{2} - x \right)} \right]dx}}$

$\small{~=~\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^5 x}{\cos^5 x + \sin^5 x} \right]dx}}$

2. Now, I + I = 2I =

$\small{\int_{0}^{\frac{\pi}{2}}{\left[\frac{\cos^5 x}{\sin^5 x + \cos^5 x} \right]dx}+\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^5 x}{\sin^5 x + \cos^5 x} \right]dx}=\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^5 x + \cos^5 x}{\sin^5 x + \cos^5 x} \right]dx}}$

$\small{~=\int_{0}^{\frac{\pi}{2}}{\left[1 \right]dx}=[x]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}}$

Therefore, $\small{I = \frac{\pi}{4}}$

Solved Example 23.111
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} \right]dx}}$
Solution:
1.1. The limits are $\small{0~\rm{and}~\frac{\pi}{2}}$. So we will apply:

$\bf{{\rm{P_4}}:\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[f(a-x) \right]dx}}$

We get:
$\small{I~=~\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} \right]dx}=\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^{\frac{3}{2}} \left(\frac{\pi}{2} - x \right)}{\sin^{\frac{3}{2}} \left(\frac{\pi}{2} - x \right) + \cos^{\frac{3}{2}} \left(\frac{\pi}{2} - x \right)} \right]dx}}$

$\small{~=~\int_{0}^{\frac{\pi}{2}}{\left[\frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} \right]dx}}$

2. Now, I + I = 2I =

$\small{\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} \right]dx}+\int_{0}^{\frac{\pi}{2}}{\left[\frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} \right]dx}=\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} \right]dx}}$

$\small{~=\int_{0}^{\frac{\pi}{2}}{\left[1 \right]dx}=[x]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}}$

Therefore, $\small{I = \frac{\pi}{4}}$


In the next section, we will see a few more solved examples.

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