Question

Improper integral question:

Answer 1

\[\large \int\limits_{0}^{\pi}\frac{\sin(x)}{\sqrt{\left| \cos(x) \right|}}dx \]

Answer 2

denominator cannot = 0, so x cannot = pi/2. So i split the integral up into 2 integrals: one from 0 to pi/2, the other pi/2 to pi. Used substitution u=cos(x) so du/dx=-sinx so sinx = = -du/dx

Answer 3

eventually came up with: -2 [ 0 - 1 ] - 2 [ 1 - 0 ] = 0

Answer 4

but i know the answer is 4 so i feel like i left out a minus somewhere

Answer 5

any help using the same substitution and method that used would be appreciated

Answer 6

Nice question. Important thing to remember is that
\[\cos x \ge 0, 0\le x\le \frac{\pi}{2}\]
\[\cos x \le 0, \frac{\pi}{2}\le x\le \pi\]
so
\[|\cos x|= \cos x\ for\ 0\le x\le \frac{\pi}{2}\]
\[|\cos x|= -\cos x\ for\ \frac{\pi}{2}\le x\le \pi\]

Answer 7

I think you missed the "-" sign for pi/2 to pi integral

Answer 8

I've done it twice tho and can't seem to find the mistake :(

Answer 9

\[\int\limits_{0}^{\pi}\frac{ \sin x }{ \sqrt{\left| \cos x \right|} }dx=\int\limits_{0}^{\pi/2}\frac{ \sin x }{ \sqrt{\cos x} }dx+\int\limits_{\pi/2}^{\pi}\frac{ \sin x }{ \sqrt{-\cos x}}dx\]

Answer 10

\[\large \int\limits_{0}^{\pi}\frac{\sin(x)}{\sqrt{\left| \cos(x) \right|}}dx\]
\[\large \int_{0} ^{\frac {\pi}{2}} \frac{\sin x}{\sqrt{\cos x}}dx+\int_{\frac{\pi}{2}} ^{\pi} \frac{\sin x}{\sqrt{-\cos x}} dx\]

For first integral substitute u= cos x , for second substitute t= -cos x
changing limits and variable
\[\int_{1} ^{0} \frac{-du}{\sqrt u} du+\int_{0} ^{-1} \frac{dt}{\sqrt t}\]
Can you do it from here? @remnant

Answer 11

Here are my workings. I know my integral bounds are wrong after the substitution and i left it as |cos(x)| but applied the || when i took the integral

Answer 12

You should have a "-" sign before the second integral

Answer 13

there is one...in about line 3...that wrong?

Answer 14

second integral will be \(\sqrt {-u}\)
u= cos x

so you'll have \[\sqrt{-\cos x}\]
now apply limits

Answer 15

don't i not need to do that as long as i keep the | | signs there?

Answer 16

After integration of the second integral you get
\[\sqrt{|u|}\]
with limits as pi/2 to pi
First of all, you have to have limits for u
which are 0 to -1
in the range 0 to -1, u is negative so, we'll have it as
\[\sqrt {-u}\]

Answer 17

i will try it again. thanks for the help :)