Q.Value of ∫₀^(π/2) cos²x dx is equal to

Split the integral where $\cos x$ changes sign on $[0,\pi]$, i.e. at $x=\pi/2$.

$\cos x \ge 0$ on $[0,\pi/2]$ and $\cos x \le 0$ on $[\pi/2,\pi]$, so $|\cos x| = \cos x$ on the first piece and $-\cos x$ on the second:

$\int_0^\pi |\cos x|\,dx = \int_0^{\pi/2}\cos x\,dx + \int_{\pi/2}^\pi (-\cos x)\,dx$

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Use the power-reduction identity $\cos^2x=\dfrac{1+\cos2x}{2}$ and integrate term by term.

Step 1. $\displaystyle\int_0^{\pi/2}\cos^2x\,dx = \int_0^{\pi/2}\frac{1+\cos2x}{2}\,dx = \frac12\Big[x+\frac{\sin2x}{2}\Big]_0^{\pi/2}$.

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