I have been playing around with some integration problems I had previously solved correctly. I attempted an approach that was a bit different on one in particular, and I am getting what seems to be a slightly incorrect answer. The original problems is:

$$ \int \frac{x+2}{\sqrt{4-x^2}}\,dx \\[30pt] $$

I previously solved the problem by splitting it up as: $$ \int \frac{x}{\sqrt{4-x^2}}\,dx + 2\int \frac{1}{\sqrt{4-x^2}}\,dx \\[30pt] $$

I ultimately found the correct answer to be:

$$ 2\cdot\sin^{-1}\Big(\frac{x}{2}\Big)-\sqrt{4-x^2}+C \\[30pt] $$

What I tried next was to solve the problem using trigonometric substitution without splitting it up. My work is as follows:

$$ \int \frac {x+2}{\sqrt{4-x^2}}\,dx = \int \frac {x+2}{\sqrt{4(1-\frac{1}{4}x^2)}}\,dx = \frac{1}{2}\int\frac{x+2}{\sqrt{1-\frac{1}{4}x^2}}\,dx \\[30pt] $$

Here I begin the trigonometric substitution with: $$ \frac{1}{2}x=\sin(\theta)\implies x=2\cdot \sin(\theta); dx=2\cdot \cos(\theta)\,d\theta \\[30pt] $$

Thus my work continues as follows:

$$ \frac{1}{2}\int\frac{x+2}{\sqrt{1-\frac{1}{4}x^2}}\,dx = \frac{1}{2}\int\frac{(2\cdot\sin(\theta)+2)}{\sqrt{1-\sin^2(\theta)}}\cdot(2\cdot \cos(\theta))\,d\theta = \\[30pt] \frac{1}{2}\int\frac{(2\cdot\sin(\theta)+2)}{\sqrt{\cos^2(\theta)}}\cdot(2\cdot \cos(\theta))\,d\theta = \frac{1}{2}\int\frac{(2\cdot\sin(\theta)+2)}{\cos(\theta)}\cdot(2\cdot \cos(\theta))\,d\theta = \\[30pt] \int (2\cdot\sin(\theta)+2)\,d\theta = 2\int (\sin(\theta)+1)\,d\theta = 2(\theta-\cos(\theta))+C = \\[30pt] 2\theta - 2\cos(\theta)+C \\[30pt] $$

Now solving for $\theta$ I get $\theta = \sin^{-1}\big(\frac{x}{2}\big)$ and solving with a right triangle I find that $\cos(\theta) = \sqrt{4-x^2} \\[30pt]$.

Thus -- substituting back in for x -- the final answer would seem to be:

$$ 2\cdot \sin^{-1}\Big( \frac{x}{2}\Big)-2\sqrt{4-x^2}+C \\[30pt] $$

But: $$ 2\cdot \sin^{-1}\Big( \frac{x}{2}\Big)-2\sqrt{4-x^2}+C \neq 2\cdot\sin^{-1}\Big(\frac{x}{2}\Big)-\sqrt{4-x^2}+C \\[30pt] $$

So what am I missing? Is there a rule necessitating that problems like this be split up before using trigonometric substitution? Have I made an error somewhere in my calculations/algebra?

  • $\begingroup$ You get $\sin \theta=x/2$, But then $\cos\theta=\sqrt{1-(x/2)^2}$ and not $\sqrt{4-x^2}$. $\endgroup$ – mickep Feb 17 '15 at 19:54

In one of the last steps:


(not $\sqrt{4-x^2}$)


Thanks to those of you that responded, I now see where I went wrong. However, I look at the mistake a bit differently. I mentioned the right triangle I used to solve one part of the problem. I do not know how (or even if it is possible) to post the image of a right triangle here. But suffice to say that I have the legs marked as x and $\sqrt{4-x^2}$. Therefore, the hypotenuse is 2. I had solved for the leg $\sqrt{4-x^2}$ and for some peculiar reason just assigned that as being equal to $\cos(\theta)$. However, $\cos(\theta)$ would actually be equal to $\frac{\sqrt{4-x^2}}{2}$ in consistency with the whole "$\frac{adjacent}{hypotenuse}$" definition of $\cos(\theta)$. With that two in the denominator, clearly it cancels the unexpected two in question and my answer here does in fact equal my previous answer.

Thank you all for your help!


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.