Exercise 2.9 - Chapter 2 Integral Calculus I 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $2.9$
Evaluate the following using properties of definite integrals:
Question 1.
$
\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} x^3 \cos ^3 x d x
$
Solution:
$
\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} x^3 \cos ^3 x d x
$
Let
$
\begin{aligned}
f(x) & =x^3 \cos ^3 x \\
f(-x) & =(-x)^3 \cos ^3(-x) \\
& =-x^3 \cos ^3 x \\
& =-f(x)
\end{aligned}
$
Here $\quad f(-x)=-f(x)$
$\therefore f(x)$ is an odd function
$
\therefore \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} x^3 \cos ^3 x d x=0
$
Question 2.
$
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 \theta d \theta
$
Solution:

$
\begin{aligned}
& \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 \theta d \theta \\
& \text { Let } f(\theta)=\sin ^2 \theta \\
& f(-\theta)=\sin ^2(-\theta)=\sin ^2 \theta \\
& =f(\theta) \\
& f(-\theta)=f(\theta) \Rightarrow f(\theta) \text { is an even function } \\
& \text { So } \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 \theta d \theta=2 \int_0^{\frac{\pi}{2}} \sin ^2 \theta d \theta \\
& =2 \int_0^{\frac{\pi}{2}} \frac{1-\cos 2 \theta}{2} d \theta\{\text { Using double angle formula }\} \\
& =\int_0^{\frac{\pi}{2}}(1-\cos 2 \theta) d \theta \\
& =\left[\theta-\frac{\sin 2 \theta}{2}\right]_0^{\frac{\pi}{2}}=\frac{\pi}{2}-\frac{\sin \pi}{2}-0+\frac{\sin 0}{2}=\frac{\pi}{2} \\
&
\end{aligned}
$
Question 3.
$
\int_{-1}^1 \log \left(\frac{2-x}{2+x}\right) d x
$

Solution:

$\int_{-1}^1 \log \left(\frac{2-x}{2+x}\right) d x$
Let $f(x)=\log \left(\frac{2-x}{2+x}\right)$
$f(-x)=\log \left(\frac{2+x}{2-x}\right)=\log (2+x)-\log (2-x)$
$=-[\log (2-x)-\log (2+x)]$
$=-\log \left(\frac{2-x}{2+x}\right)=-f(x)$
$f(-x)=-f(x) \Rightarrow f(x)$ is an odd function
So $\quad \int_{-1}^1 \log \left(\frac{2-x}{2+x}\right) d x=0$
Question 4.
$\int_0^{\frac{\pi}{2}} \frac{\sin ^7 x}{\sin ^7 x+\cos ^7 x} d x$
Solution:
$\int_0^{\frac{\pi}{2}} \frac{\sin ^7 x}{\sin ^7 x+\cos ^7 x} d x$
Let $I=\int_0^{\frac{\pi}{2}} \frac{\sin ^7 x}{\sin ^7 x+\cos ^7 x} d x$
So
$
I=\int_0^{\frac{\pi}{2}} \frac{\sin ^7\left(\frac{\pi}{2}-x\right)}{\sin ^7\left(\frac{\pi}{2}-x\right)+\cos ^7\left(\frac{\pi}{2}-x\right)}
$

$
\begin{aligned}
(1)+(2) \Rightarrow & =\int_0^{\frac{\pi}{2}} \frac{\cos ^7 x d x}{\cos ^7 x+\sin ^7 x} \\
2 I & =\int_0^{\frac{\pi}{2}}\left[\frac{\sin ^7 x}{\sin ^7 x+\cos ^7 x}+\frac{\cos ^7 x}{\cos ^7 x+\sin ^7 x}\right] d x \\
2 \mathrm{I} & =\int_0^{\frac{\pi}{2}} d x=[x]_0^{\frac{\pi}{2}}=\frac{\pi}{2} \\
\therefore \mathrm{I} & =\frac{\pi}{4}
\end{aligned}
$
Question 5.
$
\int_0^1 \log \left(\frac{1}{x}-1\right) d x
$
Solution:
$
\text { Let } \mathrm{I}=\int_0^1 \log \left(\frac{1}{x}-1\right) d x=\int_0^1 \log \left(\frac{1-x}{x}\right) d x
$
We apply $\int_0^a f(x) d x=\int_0^a f(a-x) d x$
$
\begin{aligned}
& I=\int_0^1 \log \left(\frac{1-(1-x)}{1-x}\right) d x \\
& I=\int_0^1 \log \left(\frac{x}{1-x}\right) d x
\end{aligned}
$
(1) $+(2) \Rightarrow$
$
\begin{aligned}
2 \mathrm{I} & =\int_0^1\left(\log \left(\frac{1-x}{x}\right)+\log \left(\frac{x}{1-x}\right)\right) d x \\
& =\int_0^1 \log \left(\frac{1-x}{x}\right)\left(\frac{x}{(1-x)}\right) d x\{\text { Using } \log a+\log b=\log a b\} \\
2 \mathrm{I} & =\int^1 \log 1 d x=0
\end{aligned}
$

$
\Rightarrow I=0
$
Question 6.
$
\int_0^1 \frac{x}{(1-x)^{\frac{3}{4}}} d x
$
Solution:
$
\begin{aligned}
& \int_0^1 \frac{x}{(1-x)^{\frac{3}{4}}} d x \\
& I=\int_0^1 \frac{(1-x)}{[1-(1-x)]^{\frac{3}{4}}} d x \\
& I=\int_0^1 \frac{1-x}{x^{\frac{3}{4}}} d x=\int_0^1 x^{-\frac{3}{4}} d x-\int_0^1 x^{\frac{1}{4}} d x \\
& I=\left[\frac{x^{\frac{1}{4}}}{\frac{1}{4}}\right]_0^1-\left[\frac{x^{\frac{5}{4}}}{\frac{5}{4}}\right]_0^1 \\
& =4-\frac{4}{5}=\frac{16}{5} \\
&
\end{aligned}
$