Exercise 9.6-Additional Problems - Chapter 9 - Applications of Integration - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Problems
Question 1.
Evaluate
(i) $\int_0^{\pi / 2} \sin ^7 x d x$
Solution:
$
\begin{aligned}
& \int_0^{\pi / 2} \sin ^n x d x=\frac{n-1}{n} \cdot \frac{n-3}{n-2} \cdots \frac{2}{3} \text { when ' } n \text { ' is odd } \\
& \int_0^{\pi / 2} \sin ^7 x d x=\frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3}=\frac{16}{35}
\end{aligned}
$

(iii) $\int_0^{2 \pi} \sin ^9 \frac{x}{4} d x$
Solution:

$
\begin{aligned}
\text { Put } \frac{x}{4} & =t \\
\therefore \quad d x & =4 d t \\
\int_0^{2 \pi} \sin ^9 \frac{x}{4} d x & =4 \int_0^{\pi / 2} \sin ^9 t d t \\
& =4 \cdot\left(\frac{8}{9} \cdot \frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3}\right)=\frac{512}{315} .
\end{aligned}
$

(iv) $\int_0^{\pi / 6} \cos ^7 3 x d x$
Solution:

$\begin{aligned}
\text { Put } 3 x & =t \\
\therefore 3 d x & =d t \\
d x & =\frac{1}{3} d t \\
\int_0^{\pi / 6} \cos ^7 3 x d x & =\frac{1}{3} \int_0^{\pi / 2} \cos ^7 t d t \\
& =\frac{1}{3} \cdot\left(\frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3}\right)=\frac{16}{105} .
\end{aligned}$

Question 2.
Evaluate: $\int_0^{\pi / 2} \sin ^4 x \cos ^2 x d x$
Solution:
$
\begin{aligned}
\int_0^{\pi / 2} \sin ^4 x \cos ^2 x d x & =\int_0^{\pi / 2} \sin ^4 x\left(1-\sin ^2 x\right) d x \\
& =\int_0^{\pi / 2}\left(\sin ^4 x-\sin ^6 x\right) d x=\int_0^{\pi / 2} \sin ^4 x d x-\int_0^{\pi / 2} \sin ^6 x d x \\
& =\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}-\frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}=\frac{\pi}{32} .
\end{aligned}
$