Question

another table integral

Answer 1

\[\int\limits_{0}^{1}5x \cos^-1dx\]

Answer 2

i used # 91. and i got \[\frac{ 5 }{ 4 }(\frac{ \pi }{ 2 })\] for my final answer. will type out work now

Answer 3

\[5\int\limits_{0}^{1}x \cos^-1 x dx = 5[\frac{ (2u^2-1) }{ 4 }\cos^-1u-\frac{u \sqrt{(1-u^2)} }{ 4}]\]
evaluated from 0-1 =\[5[(\frac{ 1 }{ 4 }*0-\frac{ 0 }{ 4 })-(\frac{ -1 }{ 4 }*\frac{ \pi }{ 2 }-\frac{ 0 }{ 4 })]\]=\[5[(0-(\frac{ -1 }{ 4 }*\frac{ \pi }{ 2 }-0)]\]=\[5(\frac{ 1 }{ 4 })(\frac{ \pi }{ 2 })= \frac{ 5 }{ 4 }(\frac{ \pi }{ 2 })\]

Answer 4

should my answer be \[\frac{ 5\pi }{ 8 }\]?

Answer 5

those were supposed to be \[\cos^{-1}\]

Answer 6

\[5 \int\limits_0^1 x \cos^{-1} (x) dx \\ 5[\frac{2x^2-1}{4}\cos^{-1}(x)-\frac{x \sqrt{1-x^2}}{4}]_0^1 \\5[\frac{2(1)^2-1}{4} \cos^{-1}(1)-\frac{1 \cdot \sqrt{1-1^2}}{4} ]-5[ \frac{2(0)^2-1}{4} \cos^{-1}(0)-\frac{0 \cdot \sqrt{1-0^2}}{4}] \\ 5[\frac{-1}{2}(0)-0]-5[ \frac{-1}{4}(\frac{\pi}{2})-0] \\ 0-5(\frac{-1}{4})\frac{\pi}{2}\]
yep your answer looks great to me :)

Answer 7

should the pi be separated or should it be 5pi/8

Answer 8

5pi/8 is right

( I don't know what you mean by seperate)

Answer 9

i mean is it 5/4* pi/2 or is it 5pi/8

Answer 10

\[\frac{5 \pi}{8 } \text{ is same as } \frac{5}{8} \pi\]
these are the same things
if that is what you mean

Answer 11

oh 5/4 *pi/2 is the same as 5/8 * pi or the same as 5pi/8

Answer 12

ok. good enough. thanks!

Answer 13

hate integrals still!