Question

Evaluate: ${\displaystyle {\int }_0^{\pi }x\cdot \sin x\cdot {\cos }^{2}x\cdot \text{d}x}$

Answer 1

Let I = ${\displaystyle {\int }_0^{\pi }x\cdot \sin x\cdot {\cos }^{2}x\cdot \text{d}x}$   ......(i)

∴ I = ${\displaystyle {\int }_0^{\pi }(\pi -x)\cdot \sin (\pi -x)\cdot {\left[\cos (\pi -x)\right]}^2 \text{d}x}$    ......${\displaystyle [\because {\int }_0^{\text{a}}\text{f}\left(x\right) \text{d}x={\int }_0^{\text{a}}\text{f}(\text{a}-x) \text{d}x]}$

∴ I = ${\displaystyle {\int }_0^{\pi }(\pi -x)\cdot \sin x{(-\cos x)}^2 \text{d}x}$

∴ I = ${\displaystyle {\int }_0^{\pi }(\pi -x)\cdot \sin x{\cos }^{2}x \text{d}x}$  .....(ii)

Adding (i) and (ii), we get

2I = ${\displaystyle {\int }_0^{\pi }x\cdot \sin x\cdot {\cos }^{2}x \text{d}x+{\int }_0^{\pi }(\pi -x)\cdot \sin x{\cos }^{2}x \text{d}x}$

= ${\displaystyle {\int }_0\pi (x+\pi -x)\cdot \sin x{\cos }^{2}x \text{d}x}$

∴ 2I = ${\displaystyle \pi {\int }_0^{\pi }\sin x{\cos }^{2}x \text{d}x}$

Put cos x = t

∴ − sin x dx = dt

∴ sin x dx = − dt

When x = 0, t = 1 and when x = π, t = −1

∴ 2I = ${\displaystyle \pi {\int }_1^{-1} {\text{t}}^2(-\text{dt})}$

∴ I = ${\displaystyle \frac{\pi }{2}{\int }_{-1}^1 {\text{t}}^2 \text{dt}}$  .......${\displaystyle [\because {\int }_{\text{a}}^{\text{b}}\text{f}\left(x\right) \text{d}x=-{\int }_{\text{b}}^{\text{a}} \text{f}\left(x\right) \text{d}x]}$

= ${\displaystyle \frac{\pi }{2}\times 2{\int }_0^1{\text{t}}^2 \text{dt}}$  ......[∵ t² is an even function]

= ${\displaystyle \pi {\left[\frac{{\text{t}}^2}{3}\right]}_0^1}$

= ${\displaystyle \frac{\pi }{3}(1^3-0)}$

∴ I = ${\displaystyle \frac{\pi }{3}}$