# Integrate using Trigonometric Substitutions

Evaluate the integral using trigonometric substitutions.

$$\int{ x\over \sqrt{3-2x-x^2}} \,dx$$

I am familiar with using the right triangle diagram and theta, but I do not know which terms would go on the hypotenuse and sides in this case. If you can determine which numbers or $x$-values go on the hypotenuse, adjacent, and opposite sides, I can figure out the rest, although your final answer would help me check mine. Thanks!

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Write $3-2x-x^2 =4-(x+1)^2$ first. –  David Mitra Feb 22 '12 at 3:07

The "trick" in evaluating $$\tag{1} \int{x\over\sqrt{3-2x-x^2}}\,dx$$ is to complete the square of the expression in the radicand: rewrite $3-2x-x^2$ as$$\tag{2}4-\color{maroon}{(x+1)}^2.$$

I'm not sure what this right triangle diagram you speak of is, but with the method I assume you're using, the second "trick" is to take advantage of one of the Pythagorean Identities so that the square root in $(1)$ can be taken. Looking at $(2)$, you should be reminded of $$a^2-\color{maroon}{a^2\sin^2\theta}=a^2\cos^2\theta.$$ So, one may make the substitution $$\tag{3} (x+1)=2\sin\theta.$$ Then $$4-(x+1)^2 =4-(2\sin\theta)^2= 4-4\sin^2\theta= 4\cos^2\theta$$

Also, from our substitution rule $(3)$: $dx=2\cos\theta\, d\theta$ and $x=2\sin\theta-1$.

The integral $(1)$ then becomes $$\int { 2\sin\theta-1 \over2\cos\theta}\cdot2\cos\theta\,d\theta= \int(2\sin\theta-1)\,d\theta.$$ I'll leave the rest for you...

(and now I recall the triangle business: you can label two sides using $\sin\theta=(x+1)/2$; usually, after you find the antiderivative and write back in terms of $x$, the triangle is used as an an aid to simplify your answer).

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@Izzy Ward: As shown by David Mitra, complete the square, obtaining $4-(x+1)^2$. Then if you want to use a triangle as a guide, make a right triangle with hypotenuse $2$. Label one of the "small" angles of the triangle with the label $\theta$, and let the side opposite $\theta$ be $x+1$. –  André Nicolas Feb 22 '12 at 7:20

$\int \frac{x}{\sqrt{4-(x+1)^2}}dx = \int \frac{2\sin\theta-1}{\sqrt{4-4\sin^2\theta}}(2\cos\theta)d\theta$ (using the substitution $x+1=2\sin\theta$)

$=\int\frac{2\sin\theta-1}{2\cos\theta}2\cos\theta d\theta$

$= \int (2\sin\theta-1) d\theta$

$=-2\cos\theta-\theta +C$

$=-2\left(\frac{\sqrt{3-2x-x^2}}{2}\right) - \sin^{-1}\left(\frac{x+1}{2}\right)+C$

$=-\sqrt{3-2x-x^2}- \sin^{-1}\left(\frac{x+1}{2}\right)+C$

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The last and next-to-last lines aren't equal - something funny happening with a 2. –  Gerry Myerson Feb 22 '12 at 3:36
@Gerry Myerson Thanks for pointing that out! I was experiencing LaTeX blindness - forgot to delete the 2 when I cancelled. –  Aru Ray Feb 22 '12 at 3:46