Jambmaths question:
Find the value of $\int_{0}^{\pi }{\frac{{{\cos }^{2}}\theta -1}{{{\sin }^{2}}\theta }d\theta }$
Option A:
$\pi$
Option B:
$\tfrac{\pi }{2}$
Option C:
$-\tfrac{\pi }{2}$
Option D:
$-\pi$
Jamb Maths Solution:
$\begin{align} & \int_{0}^{\pi }{\frac{{{\cos }^{2}}\theta -1}{{{\sin }^{2}}\theta }d\theta }=\int_{0}^{\pi }{\frac{{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta }d\theta } \\ & =\int_{0}^{\pi }{d\theta } \\ & =\left[ \theta \right]_{0}^{\pi }=\pi -0=\pi \end{align}$
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