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

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Question

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^{3} + x \cos x + 	an^{5} x + 1) dx$.We can split the integral as:$I = \int_{-\frac{\pi}{2}}^{\f

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$\pi$

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(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^{3} + x \cos x + 	an^{5} x + 1) dx$.We can split the integral as:$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^{3} dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \cos x dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 	an^{5} x dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 dx$.Recall the property of definite integrals: If $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) dx = 0$. If $f(x)$ is an even function,then $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$.$1$. $f_{1}(x) = x^{3}$ is an odd function,so $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^{3} dx = 0$.$2$. $f_{2}(x) = x \cos x$ is an odd function (since $(-x) \cos(-x) = -x \cos x$),so $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \cos x dx = 0$.$3$. $f_{3}(x) = 	an^{5} x$ is an odd function (since $	an^{5}(-x) = -	an^{5} x$),so $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 	an^{5} x dx = 0$.$4$. $f_{4}(x) = 1$ is an even function,so $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 dx = 2 \int_{0}^{\frac{\pi}{2}} 1 dx = 2[x]_{0}^{\frac{\pi}{2}} = 2(\frac{\pi}{2} - 0) = \pi$.Thus,$I = 0 + 0 + 0 + \pi = \pi$.Hence,the correct answer is $A$.

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^{3} + x \cos x + \tan^{5} x + 1) dx$ is

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