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$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x d x$
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2187 Upvotes
Verified Answer
The correct answer is:
2
Here $f(x)=\cos x$
$$
f(-x)=\cos (-x)=\cos x=f(x)
$$
So, $f$ is even function.
$$
\begin{aligned}
\therefore \quad & \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x d x=2 \int_0^{\frac{\pi}{2}} \cos x d x \\
& {\left[\because \int_{-a}^a f(x) d x=2 \int_0^a f(x) d x, \text { if } f \text { is even }\right] } \\
= & 2[\sin x]_0^{\pi / 2} \\
= & 2\left[\sin \frac{\pi}{2}-\sin 0\right]=2[1-0]=2
\end{aligned}
$$
$$
f(-x)=\cos (-x)=\cos x=f(x)
$$
So, $f$ is even function.
$$
\begin{aligned}
\therefore \quad & \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x d x=2 \int_0^{\frac{\pi}{2}} \cos x d x \\
& {\left[\because \int_{-a}^a f(x) d x=2 \int_0^a f(x) d x, \text { if } f \text { is even }\right] } \\
= & 2[\sin x]_0^{\pi / 2} \\
= & 2\left[\sin \frac{\pi}{2}-\sin 0\right]=2[1-0]=2
\end{aligned}
$$
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