Question

If I₁₀ =$\int_{0}^{\pi/2}{x^{10}\sin xdx}$, then I₁₀ + 90 I₈ is

• A. 10 (p/2)⁶
• B. 10 (p/2)⁹
• C. 10 (p/2)⁸
• D. 10 (p/2)⁷

Answer 1

Answer: (B)

10 (p/2)⁹

Answer 2

I₁₀ = $(–x^{10}\cos x)_{0}^{\pi/2}$+ 10$\int_{0}^{\pi/2}{x^{9}\cos xdx}$

= 10$\int_{0}^{\pi/2}{x^{9}\cos xdx}$

= 10$\left\lbrack (x^{9}\sin x)_{0}^{\pi/2} - 9\int_{0}^{\pi/2}{x^{8}\sin xdx} \right\rbrack$

= 10[(p/2)⁹ – 9I₈] Ž I₁₀ + 90I₈ = 10(p/2)⁹