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.
$\bf{{\rm{P_7~(i)}}:}$
$\bf{\int_{-a}^{a}{\left[f(x) \right]dx}~=~2 \int_0^a{\left[f(x) \right]dx}}$
♦ 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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