Evaluate the integral _0^ ^2 x d x using the result/ property. - Vidyadip

Question

Evaluate the integral $\int_0^\pi \cos ^2 x \cdot d x$ using the result/ property.

✓

Answer

$
\int_0^{2 a} f(x) d x=\int_0^a f(x) d x+\int_0^a f(2 a-x) d x
$
Let,
$
\begin{aligned}
& \mathrm{I}=\int_0^\pi \cos ^2 x \cdot d x \\
& =\int_0^{2\left(\frac{\pi}{2}\right)} \cos ^2 x \cdot d x \\
& =\int_0^{\pi / 2} \cos ^2 x \cdot d x+\int_0^{\pi / 2}\left[\cos \left(2 \frac{\pi}{2}-x\right)\right]^2 \cdot d x \\
& =\int_0^{\pi / 2} \cos ^2 x \cdot d x+\int_0^{\pi / 2} \cos ^2 x \cdot d x \\
& \because \cos (\pi-x)=-\cos x \\
& =2 \cdot \int_0^{\pi / 2} \cos ^2 x \cdot d x \\
& =\int_0^{\pi / 2}(1+\cos 2 x) \cdot d x \\
& =\left[x+\sin 2 x \cdot \frac{1}{2}\right]_0^{\pi / 2} \\
& =\left[\left(\frac{\pi}{2}+\frac{1}{2} \sin 2 \frac{\pi}{2}\right)-\left(0+\frac{1}{2} \sin 2(0)\right)\right] \\
& =\frac{\pi}{2}+0 \quad \because \sin 0=0 ; \sin \pi=0 \\
& =\frac{\pi}{2} \\
& \therefore \int_0^\pi \cos ^2 x \cdot d x=\frac{\pi}{2} \\
\end{aligned}
$

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