Mathematics - Definite Integration Question with Solution | TestHub

Let $\mathrm{I}={\int }_0^{\pi }[\cos x]dx$ $$\mathrm{I}={\int }_0^{\pi }[\cos (\pi -x)]dx={\int }_0^{\pi }[-\cos x]dx$$ On adding (1) and (2), we get $$\begin{array}{rl}& 2\mathrm{I}={\int }_0^{\pi }[\cos x]dx+{\int }_0^{\pi }[-\cos x]dx\\ & 2\mathrm{I}={\int }_0^{\pi }[\cos x]+[-\cos x]dx\\ & 2\mathrm{I}\\ & ={\int }_0^{\pi }-1dx(\because [x]+[-x]=-1\text{ if }x\notin Z)\\ & 2\mathrm{I}=-{x|}_0^{\pi }=-\pi \\ & \Rightarrow \mathrm{I}=\frac{-\pi }{2}\end{array}$$