Indefinite Integration 2 Question 4
4.. Show that $\int _0^{n \pi+v}|\sin x| d x=2 n+1-\cos v$, where $n$ is a positive integer and $0 \leq v<\pi$.
(1994, 4M)
Show Answer
Answer:
Correct Answer: 4. (4)
Solution:
- $\int _0^{n \pi+v}|\sin x| d x=\int _0^{\pi}|\sin x| d x+\int _{\pi}^{2 \pi}|\sin x| d x+\ldots$
$+\int _{(n-1) \pi}^{n \pi}|\sin x| d x+\int _{n \pi}^{n \pi+v}|\sin x| d x$
$$ =\sum _{r=1}^{n} \int _{(r-1) \pi}^{r \pi}|\sin x| d x+\int _{n \pi}^{n \pi+v}|\sin x| d x $$
Now to solve, $\int _{(r-1) \pi}^{r \pi}|\sin x| d x$, we have
$$ \begin{aligned} x & =(r-1) \pi+t \\ \Rightarrow \quad \sin x & =\sin [(r-1) \pi+t]=(-1)^{r-1} \sin t \end{aligned} $$
and when $x=(r-1) \pi, t=0$ and when
$$ x=r \pi, t=\pi $$
$\therefore \quad \int _{(r-1) \pi}^{r \pi}|\sin x| d x=\int _0^{\pi}\left|(-1)^{r-1} \sin t\right| d t$
$$ \begin{aligned} & =\int _0^{\pi}|\sin t| d t=\int _0^{\pi} \sin t d t \\ & =[-\cos t] _0^{\pi}=-\cos \pi+\cos 0=2 \end{aligned} $$
Again, $\int _{n \pi}^{n \pi+v}|\sin x| d x$, putting $x=n \pi+t$
Then, $\int _{n \pi}^{n \pi+v}|\sin x| d x=\int _0^{v}\left|(-1)^{n} \sin t\right| d t=\int _0^{v} \sin t d t$
$$ =[-\cos t] _0^{v}=-\cos v+\cos 0=1-\cos v $$
$\therefore \int _0^{n \pi+v}|\sin x| d x=\sum _{r=1}^{n} \int _{(r-1) \pi}^{r \pi} \ln \sin x\left|d x+\int _{n \pi}^{n \pi+v}\right| \sin x \mid d x$
$$ \begin{aligned} & =\sum _{r=1}^{n} 2+\int _{n \pi}^{n \pi+v}|\sin x| d x \\ & =2 n+1-\cos v \end{aligned} $$
