[amp_mcq option1=”0″ option2=”\\[\\pi – 2\\]” option3=”\\[\\pi \\]” option4=”\\[\\pi + 2\\]” correct=”option1″]<\/p>\n
The correct answer is $\\boxed{\\pi}$.<\/p>\n
To evaluate the integral, we can use the following trigonometric identity:<\/p>\n
$$\\sin^2(x) + \\cos^2(x) = 1$$<\/p>\n
We can then substitute $u = \\sin(x)$ to get:<\/p>\n
$$\\int \\sin^2(x) \\, dx = \\int \\frac{1 – \\cos^2(x)}{2} \\, dx = \\frac{x}{2} – \\frac{\\sin(2x)}{4} + C$$<\/p>\n
where $C$ is an arbitrary constant.<\/p>\n
We can then substitute back to get:<\/p>\n
$$\\int \\sin^2(x) \\, dx = \\frac{x}{2} – \\frac{\\sin(2x)}{4} + C$$<\/p>\n
Now, we can evaluate the integral $\\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} \\sin^2(x) \\, dx$. We get:<\/p>\n
$$\\begin{align} Therefore, the value of the integral $\\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} \\left( {{\\text{x}}\\cos {\\text{x}}} \\right){\\text{dx}}$ is $\\boxed{\\pi}$.<\/p>\n The other options are incorrect because they do not represent the correct value of the integral.<\/p>\n","protected":false},"excerpt":{"rendered":" [amp_mcq option1=”0″ option2=”\\[\\pi – 2\\]” option3=”\\[\\pi \\]” option4=”\\[\\pi + 2\\]” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20198","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n
\n\\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} \\sin^2(x) \\, dx &= \\left[ \\frac{x}{2} – \\frac{\\sin(2x)}{4} \\right]_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} \\
\n&= \\frac{\\pi}{2} – \\frac{\\sin(\\pi)}{4} – \\left( \\frac{-\\pi}{2} – \\frac{\\sin(-\\pi)}{4} \\right) \\
\n&= \\frac{\\pi}{2} + \\frac{\\pi}{4} \\
\n&= \\pi
\n\\end{align<\/em>}$$<\/p>\n