2021-12-07

Evaluate the integral
absolute value of $\mathrm{sin}\left(x+\pi \right)$ from 0 to $3\frac{\pi }{2}$

menerkupvd

Step 1
To evaluate the integral
${\int }_{0}^{\frac{3\pi }{2}}|\mathrm{sin}\left(x+\pi \right)|dx$
Step 2
${\int }_{0}^{\frac{3\pi }{2}}|\mathrm{sin}\left(x+\pi \right)|dx$
$={\int }_{0}^{\frac{3\pi }{2}}|\mathrm{sin}x\mathrm{cos}\pi +\mathrm{cos}x\mathrm{sin}\pi |dx$ (using $\mathrm{sin}\left(a+b\right)=\mathrm{sin}a\mathrm{cos}b+\mathrm{cos}a\mathrm{sin}b$)
$={\int }_{0}^{\frac{3\pi }{2}}|\mathrm{sin}x\left(-1\right)+\mathrm{cos}x\left(0\right)|dx$
$={\int }_{0}^{\frac{3\pi }{2}}|-\mathrm{sin}x|dx$
$={\int }_{0}^{\frac{3\pi }{2}}\mathrm{sin}xdx$
$={\left[-\mathrm{cos}x\right]}_{0}^{\frac{3\pi }{2}}$
$=\left[-\mathrm{cos}\left(\frac{3\pi }{2}\right)-\left(-cos0\right)\right]$
=[-0]-[-1]
=1
Hence, ${\int }_{0}^{\frac{3\pi }{2}}|\mathrm{sin}\left(x+\pi \right)|dx=1$.

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