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Integration by Parts

The value of ...

Question

The value of $(\int_{0}^{100 π + v} | sin x | d x) + cos V$ is $(0 \leq V \leq π)$ .

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Solution

$\int_{0}^{100 π + v} | sin x | d x = \int_{0}^{π} | sin x | d x + \int_{π}^{2 π} | sin x | d x + . . . + \int_{99 π}^{100 π} | sin x | d x + \int_{100 π}^{100 π + v} | sin x | d x$
$= 100 \sum r = 1 \int_{(r - 1) π}^{r π} | sin x | d x + \int_{100 π}^{100 π + v} | sin x | d x$
Now for $\int_{(r - 1) π}^{r π} | sin x | d x$
Let $x = (r - 1) π + t$
$\Rightarrow sin x = sin [(r - 1) π + t] = {(- 1)}^{r - 1} sin t$
When $x = (r - 1) π, t = 0$ and when $x = r π, t = π$
$∴ \int_{(r - 1) π}^{r π} | sin x | d x = \int_{0}^{π} ∣ ∣ {(- 1)}^{r - 1} l i m t ∣ ∣ d t$
$= \int_{0}^{π} | sin x | d t = \int_{0}^{π} sin t d t$
$= {[- cos t]}_{0}^{π} = - cos π + cos 0 = 2$
Again for $\int_{100 π}^{100 π + v} | sin x | d x$
Put $x = 100 π + t$
$\int_{100 π}^{100 π + v} | sin x | d x = \int_{0}^{u} {(- 1)}^{n} sin t d t = \int_{0}^{v} sin t d t$
$∴ \int_{0}^{100 π + v} | sin x | d x = 100 \sum r = 1 \int_{(r, 1) π}^{r π} | sin x | d x + \int_{100 π}^{100 π + v} | sin x | d x$
$= 100 \sum r = 1 2 + \int_{100 π}^{100 π + v} | sin x | d x = 201 - cos v$

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