Solving Integral with 2 Answers: Substituting x for sin/cos theta - Help Needed

[Appleton]

Homework Statement

$\int \frac{1}{\sqrt{1-x^{2}}} dx$

I seem to be getting two incompatible answers when I substitute x with $sin \theta$ and $cos \theta$. Could someone please help me with where I'm going wrong.

Homework Equations

The Attempt at a Solution

first attempt (the correct answer I believe):

$let\ x = sin \theta$

$dx = cos \theta\ d\theta$

$\int \frac{cos \theta}{\sqrt{1-sin^{2}\theta}} d\theta$

$= \int \frac{cos \theta}{\sqrt{cos^{2}\theta}} d\theta$

$= \int \frac{cos \theta}{{cos\theta}} d\theta$

$= \int 1\ d\theta$

$= \theta$

$x = sin \theta$

$\theta = arcsin x$


second attempt (the wrong answer I believe)

$let\ x = cos \theta$

$dx = -sin \theta\ d\theta$

$\int \frac{-sin \theta}{\sqrt{1-cos^{2}\theta}} d\theta$

$= \int \frac{-sin \theta}{\sqrt{sin^{2}\theta}} d\theta$

$= \int \frac{-sin \theta}{{sin\theta}} d\theta$

$= \int -1\ d\theta$

$= -\theta$

$x = cos \theta$

$\theta = -arccos x$

 

[sandyp]

Its simple...we know that d/dx arc sinx = (1-x^2)^-1/2
d/dx arc cos x = -(1-x^2)^-1/2 (notice the minus sign)
as integral is antiderivative of a function you are getting the same values
yes integral arc sinx=(minus)integral arc cos x..Hope your doubt is clarified

 

[ShayanJ]

In fact $\int \frac{1}{\sqrt{1-x^2}}dx=\sin^{-1}{x}+C_1 \ and \ \int \frac{1}{\sqrt{1-x^2}}dx=-\cos^{-1}{x}+C_2$
If you take a look at the plots of the functions $\sin^{-1}{x}$ and $\cos^{-1}{x}$,you can see that you can get one of them by multiplying the other by a minus sign and then adding a constant(or vice versa!)...and you can see that you have the negative sign and also the constant in the results of the integral!

 

[Appleton]

Thanks for clearing that up for me.

 

[Mark44]

No one has mentioned this. There's a trig identity that can shed some light here.

$sin^{-1}(x) + cos^{-1}(x) = \pi/2$
The two functions differ by a constant. If an indefinite integral produces two different antiderivatives, those functions can differ by at most a constant.