# Are there restrictions I have forgotten for Integration by Trigonometric Substitution or am I making some other mistake?

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?

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

$\cos\theta=\cos(\arcsin(\frac{x}{2}))=\sqrt{1-\frac{x^2}{4}}$
(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.