Exercise 9.6 - Chapter 9 Applications of Integration 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $9.6$
Question 1.
(i) $\int_0^{\frac{\pi}{2}} \sin ^{10} x d x$
Solution:
$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \sin ^{10} x d x & =\frac{9}{10} \times \frac{7}{8} \times \frac{5}{6} \times \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2} \\
& =\frac{63 \pi}{512}
\end{aligned}
$

(ii)
$
\int_0^{\frac{\pi}{2}} \cos ^7 x d x
$
Solution:
$
\int_0^{\frac{\pi}{2}} \cos ^7 x d x=\frac{6}{7} \times \frac{4}{5} \times \frac{2}{3}=\frac{16}{35}
$

$\text { (iii) } \int_0^{\frac{\pi}{4}} \sin ^6 2 x d x$

Solution

$
\begin{aligned}
& \int_0^{\frac{\pi}{4}} \sin ^6 2 x d x \\
& \text { Put } t=2 x
\end{aligned}
$
Differentiate with respect to $x^{\prime}$
$
\begin{aligned}
d t & =2 d x \\
\frac{d t}{2} & =d x
\end{aligned}
$
Substitute (2) and (3) in (1)
$
\begin{aligned}
(1) \Rightarrow \int_0^{\frac{\pi}{2}} & \sin ^6 t \frac{d t}{2} \\
& =\frac{1}{2} \int_0^{\frac{\pi}{2}} \sin ^6 t d t \\
& =\frac{1}{2}\left[\frac{5}{6} \times \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}\right]=\frac{5 \pi}{64} .
\end{aligned}
$

(iv) $\int_0^{\frac{\pi}{6}} \sin ^5 3 x d x$
Solution:

$
\text { Put } \begin{aligned}
t & =3 x \\
& =\frac{1}{3} \int_0^{\frac{\pi}{2}} \sin ^5 t d t \\
& =\frac{1}{3}\left[\frac{4}{5} \times \frac{2}{3}\right]=\frac{8}{45} .
\end{aligned}
$

(v) $\int_0^{\frac{\pi}{2}} \sin ^2 x \cos ^4 x d x$
Solution:
$
\begin{aligned}
& \int_0^{\frac{\pi}{2}} \sin ^2 x \cos ^4 x d x \\
& =\int_0^{\frac{\pi}{2}}\left(1-\cos ^2 x\right) \cos ^4 x d x=\int_0^{\frac{\pi}{2}}\left(\cos ^4 x-\cos ^6 x\right) d x \\
& =\int_0^{\frac{\pi}{2}} \cos ^4 x d x-\int_0^{\frac{\pi}{2}} \cos ^6 x d x \\
& =\left(\frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}\right)-\left(\frac{5}{6} \times \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}\right) \\
& =\frac{3 \pi}{16}-\frac{5 \pi}{32}=\frac{\pi}{16}\left[3-\frac{5}{2}\right] \\
& =\frac{\pi}{16}\left[\frac{6-5}{2}\right]=\frac{\pi}{16}\left(\frac{1}{2}\right)=\frac{\pi}{32} . \\
&
\end{aligned}
$

(vi) $\int_0^{2 \pi} \sin ^7 \frac{x}{4} d x$
Solution:

$
\text { Put } \begin{aligned}
t & =\frac{x}{4} \\
& =4 \int_0^{\frac{\pi}{2}} \sin ^7 t d t \\
& =4\left[\frac{6}{7} \times \frac{4}{5} \times \frac{2}{3}\right]=\frac{64}{35} .
\end{aligned}
$

(vii) $\int_0^{\frac{\pi}{2}} \sin ^3 \theta \cos ^5 \theta d \theta$
Solution:
We know that,
$
\int_0^{\frac{\pi}{2}} \sin ^m x \cos ^n x d x=\frac{[(m-1)(m-3) \ldots 2.1][(n-1)(n-3) \ldots 2.1]}{(m+n)(m+n-2)(m+n-4) \ldots \ldots . .}
$
Here $m=3, n=5$
$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \sin ^3 \theta \cos ^5 \theta d \theta & =\frac{[(3-1)][(5-1)(5-3)]}{(3+5)(3+5-2)(3+5-4)(3+5-6)} \\
& =\frac{(2)(4)(2)}{(8)(6)(4)(2)}=\frac{1}{24} .
\end{aligned}
$
(viii) $\int_0^1 x^2(1-x)^3 d x$
Solution:
We know that,
$
\begin{aligned}
& \int_0^1 x^m(1-x)^n d x=\frac{m ! \times n !}{(m+n+1) !} \\
& \int_0^1 x^2(1-x)^3 d x=\frac{2 ! \times 3 !}{(2+3+1) !}=\frac{2 \times 6}{1 \times 2 \times 3 \times 4 \times 5 \times 6}=\frac{1}{60} .
\end{aligned}
$