{"id":20188,"date":"2024-04-15T05:49:33","date_gmt":"2024-04-15T05:49:33","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20188"},"modified":"2024-04-15T05:49:33","modified_gmt":"2024-04-15T05:49:33","slug":"the-value-of-the-integral-intlimits_02-fracleft-textx-1-right2sin-left-textx-1-rightleft-textx-1-right2-cos-left","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-value-of-the-integral-intlimits_02-fracleft-textx-1-right2sin-left-textx-1-rightleft-textx-1-right2-cos-left\/","title":{"rendered":"The value of the integral \\[\\int\\limits_0^2 {\\frac{{{{\\left( {{\\text{x}} – 1} \\right)}^2}\\sin \\left( {{\\text{x}} – 1} \\right)}}{{{{\\left( {{\\text{x}} – 1} \\right)}^2} + \\cos \\left( {{\\text{x}} – 1} \\right)}}} {\\text{dx}}\\] is A. 3 B. 0 C. -1 D. -2"},"content":{"rendered":"

[amp_mcq option1=”3″ option2=”0″ option3=”-1″ option4=”-2″ correct=”option1″]<\/p>\n

The correct answer is $\\boxed{\\text{B) }0}$.<\/p>\n

To solve this integral, we can use the following identity:<\/p>\n

$$\\int \\frac{\\sin(x)}{1+\\cos(x)} \\, dx = \\text{Si}(x) + C$$<\/p>\n

where $\\text{Si}(x)$ is the sine integral function.<\/p>\n

Substituting $x = x-1$, we get:<\/p>\n

$$\\int \\frac{\\sin(x-1)}{1+\\cos(x-1)} \\, dx = \\text{Si}(x-1) + C$$<\/p>\n

The limits of integration are $0$ and $2$. Substituting these in, we get:<\/p>\n

$$\\text{Si}(2) – \\text{Si}(-1) = 0$$<\/p>\n

Therefore, the value of the integral is $\\boxed{\\text{B) }0}$.<\/p>\n

Here is a brief explanation of each option:<\/p>\n