Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
2  
Write $3-2x-x^2 =4-(x+1)^2$ first. –  David Mitra Feb 22 '12 at 3:07
add comment

2 Answers 2

up vote 5 down vote accepted

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).

share|improve this answer
    
@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
add comment

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

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

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

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

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

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

share|improve this answer
3  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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