9. evaluate the integrals\n(i) $\\int_{-2}^{2} \\frac{x}{\\sqrt{x^{2}+1}} d x$\n(ii) $\\int_{-\\pi /…

Explanation:

Step1: Check for odd - function property

Let (f(x)=\frac{x}{\sqrt{x^{2}+1}}). Then (f(-x)=\frac{-x}{\sqrt{(-x)^{2}+1}}=-\frac{x}{\sqrt{x^{2}+1}}=-f(x)). For a function (y = f(x)) defined on ([-a,a]), if (f(-x)=-f(x)) (odd function), then (\int_{-a}^{a}f(x)dx = 0). Here (a = 2) and (f(x)=\frac{x}{\sqrt{x^{2}+1}}) is an odd function.

Step2: Evaluate the second integral

Let (I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(4x\cos x+x^{2}\tan x + 2)dx=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}4x\cos xdx+\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}x^{2}\tan xdx+\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}2dx)

• For (y = 4x\cos x), let (f(x)=4x\cos x). Then (f(-x)=4(-x)\cos(-x)=-4x\cos x=-f(x)). So (\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}4x\cos xdx = 0)
• For (y=x^{2}\tan x), let (g(x)=x^{2}\tan x). Then (g(-x)=(-x)^{2}\tan(-x)=-x^{2}\tan x=-g(x)). So (\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}x^{2}\tan xdx = 0)
• For (y = 2), (\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}2dx=2\left[x\right]{-\frac{\pi}{4}}^{\frac{\pi}{4}}) Using the fundamental theorem of calculus (\int{a}^{b}kdx=k(b - a)) (where (k = 2), (a=-\frac{\pi}{4}), (b=\frac{\pi}{4})) (2\left(\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right)=2\times\frac{\pi}{2}=\pi)

Answer:

(i) (0) (ii) (\pi)