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.
Solved Example 23.112
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right]dx}}$
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:
$\small{I~=~\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\right]dx}~=~\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sqrt{\sin \left(\frac{\pi}{2} - x \right)}}{\sqrt{\sin \left(\frac{\pi}{2} - x \right)} + \sqrt{\cos \left(\frac{\pi}{2} - x \right)}} \right]dx} }$
$\small{~=~\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}\right]dx}}$
2. Now, I + I = 2I =
$\small{\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right]dx}+\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \right]dx}=\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sqrt{\sin x}~+~\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right]dx}}$
$\small{~=\int_{0}^{\frac{\pi}{2}}{\left[1 \right]dx}=[x]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}}$
Therefore, $\small{I = \frac{\pi}{4}}$
Solved Example 23.113
Evaluate $\small{\int_{0}^{a}{\left[\frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} \right]dx}}$
Solution:
1. The limits are $\small{0~\rm{and}~a}$. 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:
$\small{I~=~\int_{0}^{a}{\left[\frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}}\right]dx}~=~\int_{0}^{a}{\left[\frac{\sqrt{a-x}}{\sqrt{a-x} ~+~ \sqrt{a-(a-x)}}\right]dx} }$
$\small{~=~\int_{0}^{a}{\left[\frac{\sqrt{a-x}}{\sqrt{a-x} ~+~ \sqrt{x}}\right]dx}}$
2. Now, I + I = 2I =
$\small{\int_{0}^{a}{\left[\frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}}\right]dx}+\int_{0}^{a}{\left[\frac{\sqrt{a-x}}{\sqrt{a-x} ~+~ \sqrt{x}}\right]dx}=\int_{0}^{a}{\left[\frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x} + \sqrt{a-x}}\right]dx}}$
$\small{~=\int_{0}^{a}{\left[1 \right]dx}=[x]_{0}^{a}=a}$
Therefore, $\small{I = \frac{a}{2}}$
Solved Example 23.114
Evaluate $\small{\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}{\left[\sin^7 x \right]dx}}$
Solution:
1. The limits are $\small{\frac{-\pi}{2}~\rm{and}~\frac{\pi}{2}}$.
So we check whether $\small{f(x)=\sin^7 x}$ is even or odd.
(a) $\small{f(-x)\,=\,\sin^7 (-x)\,=\,\left(\sin(-x) \right)^7}$
$\small{\,=\,\left(-\sin(x) \right)^7\,=\,-\sin^7 x}$
$\small{\because \sin(-x) = -\sin x}$
(b) $\small{-f(x)\,=\,-\sin^7 (x)}$
(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: $\small{\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}{\left[\sin^7 x \right]dx}~=~0}$
Solved Example 23.115
Evaluate $\small{\int_{0}^{2\pi}{\left[\cos^5 x \right]dx}}$
Solution:
1. The limits are 0 and $\small{2 \pi}$.
So we can apply:
$\bf{{\rm{P_6}}:}$
$\bf{\int_0^{2a}{\left[f(x) \right]dx}~=~2 \int_0^a{\left[f(x) \right]dx}}$
♦ If $\small{f(2a - x)~=~f(x)}$
$\bf{\int_0^{2a}{\left[f(x) \right]dx}~=~0}$
♦ If $\small{f(2a - x)~=~-f(x)}$
Here, 2a = $\small{2 \pi}$
So $\small{f(2a-x) = \cos^5(2 \pi - x)=\left(\cos(2\pi -x) \right)^5 =\left(\cos(x) \right)^5 = \cos^5 x = f(x)}$
So we can apply P6 (i). We get:
$\small{\int_{0}^{2\pi}{\left[\cos^5 x \right]dx}=2\int_{0}^{\pi}{\left[\cos^5 x \right]dx}~=~2I}$
$\bf{{\rm{P_4}}:\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[f(a-x) \right]dx}}$
We get:
$\small{I = \int_{0}^{\pi}{\left[\cos^5 x \right]dx}=\int_{0}^{\pi}{\left[\cos^5 (\pi - x) \right]dx} }$
$\small{\int_{0}^{\pi}{\left[(-\cos x)^5 \right]dx} =\int_{0}^{\pi}{\left[(-1)\cos^5x\right]dx}=(-1)\int_{0}^{\pi}{\left[\cos^5x\right]dx}=(-1)I}$
[$\small{\because~\cos(\pi - x)=-\cos x}$]
• Then $\small{2I = 0}$
3. Therefore, from the result in (1), we get:
$\small{\int_{0}^{2\pi}{\left[\cos^5 x \right]dx}~=~2I~=~0}$
Solved Example 23.116
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[2 \log \sin(x) - \log \sin(2x) \right]dx}}$
Solution:
1. Let us rearrange the given function:
$\small{f(x)=2 \log \sin(x) - \log \sin(2x)= \frac{}{} \log \sin^2(x) - \log \sin(2x)}$
$\small{= \log\left(\frac{\sin^2(x)}{\sin(2x)} \right)= \log\left(\frac{\sin^2(x)}{2 \sin x \cos x} \right)=\log(\sin x)-\log(\cos x)-\log 2}$
2. So we can write:
$\small{I = \int_{0}^{\frac{\pi}{2}}{\left[2 \log \sin(x) - \log \sin(2x) \right]dx}=\int_{0}^{\frac{\pi}{2}}{\left[\log(\sin x)-\log(\cos x)-\log 2\right]dx}}$
$\small{\Rightarrow I =\int_{0}^{\frac{\pi}{2}}{\left[\log(\sin x)-\log(\cos x)\right]dx}+(-1)\int_{0}^{\frac{\pi}{2}}{\left[\log 2\right]dx}}$
3. Applying P4, we get:
$\small{I=\int_{0}^{\frac{\pi}{2}}{\left[\log(\sin (\frac{\pi}{2}-x))-\log(\cos (\frac{\pi}{2}-x))\right]dx}+(-1)\int_{0}^{\frac{\pi}{2}}{\left[\log 2\right]dx}}$
$\small{\Rightarrow I=\int_{0}^{\frac{\pi}{2}}{\left[\log(\cos x)-\log(\sin x)\right]dx}+(-1)\int_{0}^{\frac{\pi}{2}}{\left[\log 2\right]dx}}$
$\small{\Rightarrow I=(-1)\int_{0}^{\frac{\pi}{2}}{\left[\log(\sin x)-\log(\cos x)\right]dx}+(-1)\int_{0}^{\frac{\pi}{2}}{\left[\log 2\right]dx}}$
4. Let us compare the results from (2) and (3):
• From (2), we have:
$\small{I =\int_{0}^{\frac{\pi}{2}}{\left[\log(\sin x)-\log(\cos x)\right]dx}+(-1)\int_{0}^{\frac{\pi}{2}}{\left[\log 2\right]dx}}$
• From (3), we have:
$\small{I=(-1)\int_{0}^{\frac{\pi}{2}}{\left[\log(\sin x)-\log(\cos x)\right]dx}+(-1)\int_{0}^{\frac{\pi}{2}}{\left[\log 2\right]dx}}$
5. Adding the two results, we get:
$\small{2I=(-2)\int_{0}^{\frac{\pi}{2}}{\left[\log 2\right]dx}}$
$\small{\Rightarrow I=(-1)\int_{0}^{\frac{\pi}{2}}{\left[\log 2\right]dx}=(-1)(\log 2)[x]_0^{\frac{\pi}{2}}}$
$\small{=(-1)(\log 2)\left[\frac{\pi}{2} \right]=\frac{\pi}{2} \left[\log\left( \frac{1}{2}\right) \right]}$
Solved Example 23.117
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\log \sin(x) \right]dx}}$
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:
$\small{I~=~\int_{0}^{\frac{\pi}{2}}{\left[\log \sin(x)\right]dx}~=~\int_{0}^{\frac{\pi}{2}}{\left[\log \sin \left(\frac{\pi}{2} - x \right) \right]dx} }$
$\small{~=~\int_{0}^{\frac{\pi}{2}}{\left[\log \cos(x)\right]dx}}$
2. Now, I + I = 2I =
$\small{\int_{0}^{\frac{\pi}{2}}{\left[\log \sin(x) \right]dx}+\int_{0}^{\frac{\pi}{2}}{\left[\log \cos(x) \right]dx}=\int_{0}^{\frac{\pi}{2}}{\left[\log \sin(x) + \log \cos(x) \right]dx}}$
$\small{~=\int_{0}^{\frac{\pi}{2}}{\Big[\log \left[\sin(x)\,\cos x \right] \Big]dx}}$
3. Let us add and subtract "log 2". We get:
$\small{2I~=\int_{0}^{\frac{\pi}{2}}{\Big[\log \left[\sin(x)\,\cos x \right]~+~\log 2~-~\log 2 \Big]dx}}$
$\small{\Rightarrow 2I~=~\int_{0}^{\frac{\pi}{2}}{\Big[\log \left[2 \sin(x)\,\cos x \right]~-~\log 2 \Big]dx}~=~\int_{0}^{\frac{\pi}{2}}{\Big[\log \left[\sin(2x) \right]~-~\log 2 \Big]dx}}$
$\small{\Rightarrow 2I~=~\int_{0}^{\frac{\pi}{2}}{\Big[\log \left[\sin(2x) \right] \Big]dx}~-~\int_{0}^{\frac{\pi}{2}}{\Big[\log 2 \Big]dx}~=~I_1~-~I_2}$
4. Now we will evaluate $\small{I_1}$:
We have: $\small{I_1~=~\int_{0}^{\frac{\pi}{2}}{\Big[\log \left[\sin(2x) \right] \Big]dx}}$
• Put u = 2x
• Then $\small{\frac{du}{dx}~=~2 ~\Rightarrow 2dx = du}$
Also, When x approach zero, u approach zero
And when x approach $\small{\frac{\pi}{2}}$, u approach $\small{\pi}$
Then $\small{I_1~=~\int_{0}^{\frac{\pi}{2}}{\Big[\left(\frac{2}{2} \right)\log \left[\sin(2x) \right] \Big]dx}~=~\int_{0}^{\pi}{\Big[\left(\frac{1}{2} \right)\log \left[\sin(u) \right] \Big]du}}$
$\small{\Rightarrow I_1~=~\left(\frac{1}{2} \right) \int_{0}^{\pi}{\Big[\log \left[\sin(u) \right] \Big]du}}$
5. Let us apply:
$\bf{{\rm{P_6}}:}$
$\bf{\int_0^{2a}{\left[f(x) \right]dx}~=~2 \int_0^a{\left[f(x) \right]dx}}$
♦ If $\small{f(2a - x)~=~f(x)}$
$\bf{\int_0^{2a}{\left[f(x) \right]dx}~=~0}$
♦ If $\small{f(2a - x)~=~-f(x)}$
• Here assume that, 2a = $\small{\pi}$
• Given $\small{f(u)=\log[\sin(u)]}$
• $\small{f(2a-u)=\log[\sin(\pi-u)]~=~\log[\sin (u)]~=~f(u)}$
So we can apply P6 (i). We get:
$\small{I_1~=~\left(\frac{1}{2} \right) \int_{0}^{\pi}{\Big[\log \left[\sin(u) \right] \Big]du}}$
$\small{\int_{0}^{\pi}{\Big[\log \left[\sin(u) \right] \Big]du}~=~2\int_{0}^{\frac{\pi}{2}}{\Big[\log \left[\sin(u) \right] \Big]du}}$
6. Then from (4), we get:
$\small{I_1~=~\left(\frac{1}{2} \right) \int_{0}^{\pi}{\Big[\log \left[\sin(u) \right] \Big]du}~=~\left(\frac{1}{2} \right)(2) \int_{0}^{\frac{\pi}{2}}{\Big[\log \left[\sin(u) \right] \Big]du}}$
$\small{\Rightarrow I_1~=~ \int_{0}^{\frac{\pi}{2}}{\Big[\log \left[\sin(u) \right] \Big]du}}$
7. Applying P0 to the result in (6), we can write:
$\small{I_1~=~ \int_{0}^{\frac{\pi}{2}}{\Big[\log \left[\sin(x) \right] \Big]dx}}$
$\small{\Rightarrow I_1~=~ I}$
8. Now, based on the result in (3), we can write:
$\small{2I~=~I~-~I_2}$
$\small{\Rightarrow I~=~(-1)I_2}$
$\small{\Rightarrow I~=~(-1)\int_{0}^{\frac{\pi}{2}}{\Big[\log 2 \Big]dx}}$
$\small{\Rightarrow I~=~(-1){\Big[\log 2\left(\frac{\pi}{2} \right) \Big]dx}~=~\frac{-\pi\,\log 2}{2}}$
Solved Example 23.118
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\log\left(\frac{4 + 3 \sin x}{4 + 3 \cos x} \right) \right]dx}}$
Solution:
1. The limits are $\small{0~\rm{and}~a}$. 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:
$\small{I~=~\int_{0}^{\frac{\pi}{2}}{\left[\log\left(\frac{4 + 3 \sin x}{4 + 3 \cos x} \right) \right]dx}~=~\int_{0}^{\frac{\pi}{2}}{\left[\log\left(4 + 3 \sin x \right)~-~\log\left(4 + 3 \cos x \right) \right]dx} }$
$\small{~=~\int_{0}^{\frac{\pi}{2}}{\left[\log\left(4 + 3 \sin \left(\frac{\pi}{2} - x \right) \right)~-~\log\left(4 + 3 \cos \left(\frac{\pi}{2} - x \right) \right) \right]dx} }$
$\small{~=~\int_{0}^{\frac{\pi}{2}}{\left[\log\left(4 + 3 \cos x \right)~-~\log\left(4 + 3 \sin x \right) \right]dx} }$
$\small{~=~(-1)\int_{0}^{\frac{\pi}{2}}{\left[\log\left(4 + 3 \sin x \right)~-~\log\left(4 + 3 \cos x \right) \right]dx} }$
$\small{~=~(-1)I }$
2. Therefore, 2I = 0, Which gives I = 0
The link below gives a few more solved examples:
Exercise 23.11
In the next section, we will see some miscellaneous solved examples.
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