# substitution integration question

I want to integrate $\int \sqrt{1 - x^2} dx$.

When I substitute $x = \sin θ$ , I get the right answer. ( $\cos^2\theta$ before integration)

But when I substitute $x = \cos θ$ , I don't get the right answer. ( $-\sin^2\theta$ before integration).

What step is wrong here? If I proceed like this, don't I end up with a wrong answer?

• I get the same result with both substitutions. Can you show us your work so we can see where the problem is? – 5xum Apr 15 '15 at 8:41
• @5xum I'm guessing he/she's having trouble with the derivative of $\cos(\theta)$ having a negative term, and the $\sin(\theta)$ doesn't. – nathan.j.mcdougall Apr 15 '15 at 8:47
• @nathan.j.mcdougall Possibly. But unless he shows his work, we cannot really help him. – 5xum Apr 15 '15 at 8:47
• When I substitute x = sinθ, the equivalent integration function becomes cosθ (cos^2 θ)^0.5. The other substitution yields -sinθ (sin^2 θ)^0.5. Am I wrong in this step? – lgj Apr 15 '15 at 8:54
• Hint: $\sqrt{1-\sin^2 \theta}= |\cos \theta| \ne \cos \theta$. and the same for $\sqrt{1-\cos^2 \theta}= |\sin \theta| \ne \sin \theta$, so the integration requaire a bit more attention. – Emilio Novati Apr 15 '15 at 10:37

The short answer is they become the same answer once you back-substitute $\theta$ for $x$.

It makes more sense if you actually do the integrals.

\begin{align} \int \cos^2 \theta \,d\theta &= \frac{1}{2}\int (1 +\cos 2\theta)\,d\theta \\&= \frac{1}{2}\theta + \frac{1}{4}\sin 2\theta + C \\&= \frac{1}{2}\theta + \frac{1}{2}\sin\theta\cos\theta + C \\&= \frac{1}{2}\arcsin x + \frac{1}{2}x\sqrt{1-x^2} + C \end{align}

\begin{align} \int -\sin^2 \theta \,d\theta &= -\frac{1}{2}\int (1 -\cos 2\theta)\,d\theta \\&= -\frac{1}{2}\theta + \frac{1}{4}\sin 2\theta + C \\&= -\frac{1}{2}\theta + \frac{1}{2}\sin\theta\cos\theta + C \\&= -\frac{1}{2}\arccos x + \frac{1}{2}x\sqrt{1-x^2} + C \end{align}

The key here is that $\arcsin x$ and $-\arccos x$ differ by a constant

$$\arcsin x + C_1 = -\arccos x + \frac{\pi}{2} + C_1 = -\arccos x + C_2$$

What this means is when you take the derivative, the constant goes away, leaving the same function, so the 2 answers you get are equivalent.