# Thread: integration by parts/ inverse trig function

1. ## integration by parts/ inverse trig function

here is the problem
evaluate the Integral
arcsin(x)/ sqrt(1+x^2)
the limits are 0 and .5

I've used integration by part and came up with the equation of
integral = arcsin^2(x) - arctanx
then i put in the limits and got 873.44 this is when the calc is set to degrees
but i have a feeling im doing this wrong
please i need help i have a mid term later today

here is some of my work
u= arcsinx dv=1/sqrt(1+x^2)
du= 1/sqrt(1+x^2) v=arcsinx
arcsinx/ sqrt(1+x^2)= arcsinx(arcsinx)- intg (1/sqrt(1+x^2)^2

this is then simplified to
arcsinx^2-arctanx
then by plugging in .5
it becomes (pi/6)^2- arctan(.5)
which comes out to -.189

2. Originally Posted by Lone21
here is the problem
evaluate the Integral
arcsin(x)/ sqrt(1+x^2)
the limits are 0 and .5

I've used integration by part and came up with the equation of
integral = arcsin^2(x) - arctanx
then i put in the limits and got 873.44 this is when the calc is set to degrees
but i have a feeling im doing this wrong
please i need help i have a mid term later today

here is some of my work
u= arcsinx dv=1/sqrt(1+x^2)
du= 1/sqrt(1+x^2) v=arcsinx
arcsinx/ sqrt(1+x^2)= arcsinx(arcsinx)- intg (1/sqrt(1+x^2)^2

this is then simplified to
arcsinx^2-arctanx
then by plugging in .5
it becomes (pi/6)^2- arctan(.5)
which comes out to -.189
If this is indeed your integral $\displaystyle \int_0^{\frac{1}{2}}\frac{\arcsin(x)}{\sqrt{1+x^2} }$, then you cannot do this by elementary terms. Are you sure it wasn't

$\displaystyle \int_0^{\frac{1}{2}}\frac{\arcsin(x)}{\sqrt{1-x^2}}$

3. Originally Posted by Mathstud28
If this is indeed your integral $\displaystyle \int_0^{\frac{1}{2}}\frac{\arcsin(x)}{\sqrt{1+x^2} }$, then you cannot do this by elementary terms. Are you sure it wasn't

$\displaystyle \int_0^{\frac{1}{2}}\frac{\arcsin(x)}{\sqrt{1-x^2}}$
yes thats the correct question
i think i was on the right path but just had the wrong equation
if you could evaluate how its done that would be great

4. Originally Posted by Lone21
yes thats the correct question
i think i was on the right path but just had the wrong equation
if you could evaluate how its done that would be great
Do you still agree with your work? $\displaystyle \frac{d}{dx}\bigg[\arcsin(x)\bigg]=\frac{1}{\sqrt{1{\color{red}-}x^2}}$