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?

  • 5
    $\begingroup$ I get the same result with both substitutions. Can you show us your work so we can see where the problem is? $\endgroup$
    – 5xum
    Apr 15 '15 at 8:41
  • $\begingroup$ @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. $\endgroup$ Apr 15 '15 at 8:47
  • 3
    $\begingroup$ @nathan.j.mcdougall Possibly. But unless he shows his work, we cannot really help him. $\endgroup$
    – 5xum
    Apr 15 '15 at 8:47
  • $\begingroup$ 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? $\endgroup$
    – lgj
    Apr 15 '15 at 8:54
  • 1
    $\begingroup$ 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. $\endgroup$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.