Question

99 POINT QUESTION, AND BRAINLIEST!!!

Answer 1

$\displaystyle\int_{x=1}^{x=14}\frac{1-\cos4x}{4x-\sin4x}\,\mathrm dx$

First, replace $y=4x$ and $\mathrm du=4\,\mathrm dx$. Then you have

$\displaystyle\frac14\int_{y=4}^{y=56}\frac{1-\cos y}{y-\sin y}\,\mathrm dy$

If we take $z=y-\sin y$, then immediately we see a differential in the numerator in the form of $\mathrm dz=(1-\cos y)\,\mathrm dy$. So the integral becomes

$\displaystyle\frac14\int_{z=4-\sin4}^{z=56-\sin56}\frac{\mathrm dz}z$

and evaluates to

$\dfrac14\ln|z|\bigg|_{z=4-\sin4}^{z=56-\sin56}=\dfrac14\left(\ln|56-\sin56|-\ln|4-\sin4|\right)=\dfrac14\ln\left|\dfrac{56-\sin56}{4-\sin4}\right|$

making (A) the correct answer.

Answer 2

This question is quite difficult but I've managed to solve it.

The correct answer is A.