I am trying to solve the following integral with trig substitutions. However, I get a different answer for two substitutions that should yield the same result. $$\int\frac{dx}{\sqrt{4-9x^2}}$$
For the first trig sub, I set $9x^2 = 4\cos^2\theta$. This simplifies to: $x = \frac{2}{3}\cos\theta$, and $dx = -\frac{2}{3}\sin\theta$. Substituting in, I get: $$\int\frac{-2\sin\theta}{6\sin\theta} = -\frac{\theta}{3} = -\frac{1}{3}\,\cos^{-1}\left(\frac{3x}{2}\right)+C \tag{1}$$
For the second trig sub, I set $9x^2 = 4\sin^2\theta$. This simplifies to: $x = \frac{2}{3}\sin\theta$, and $dx = \frac{2}{3}\cos\theta$. Substituting in, I get: $$\int\frac{2\cos\theta}{6\cos\theta} = \frac{\theta}{3} = \frac{1}{3}\,\sin^{-1}\left(\frac{3x}{2}\right)+C \tag{2}$$
My question is:
Why do these two trig substitutions yield different results graphically? Shouldn't they result in the same graph?
Since $(-\arcsin)'(x)=\arccos'(x)=-\dfrac1{\sqrt{1-x^2}}$, you got twice the same thing.