The value of int_{-frac{pi}{2}}^{frac{pi}{2}} | vedclass

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Question

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(x^3+\cos x+	an^5 x) dx$. We can split this into three integrals: $I = \int_{-\frac{\pi}{2}}^{\frac

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$2$

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(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(x^3+\cos x+	an^5 x) dx$. We can split this into three integrals: $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^3 dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 	an^5 x dx$. Recall the property of definite integrals: $\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is an odd function,and $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if $f(x)$ is an even function. For $f_1(x) = x^3$,$f_1(-x) = (-x)^3 = -x^3 = -f_1(x)$,so it is an odd function. Thus,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^3 dx = 0$. For $f_2(x) = \cos x$,$f_2(-x) = \cos(-x) = \cos x = f_2(x)$,so it is an even function. Thus,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x dx = 2 \int_{0}^{\frac{\pi}{2}} \cos x dx = 2[\sin x]_{0}^{\frac{\pi}{2}} = 2(1-0) = 2$. For $f_3(x) = 	an^5 x$,$f_3(-x) = 	an^5(-x) = -	an^5 x = -f_3(x)$,so it is an odd function. Thus,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 	an^5 x dx = 0$. Therefore,$I = 0 + 2 + 0 = 2$.

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(x^3+\cos x+\tan^5 x) dx$ is equal to . . . . . . .

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