# Math Help - trig substitutions

1. ## trig substitutions

1. integral (sqrt (x^2-a^2))/x^4 dx

i can get up to integral (a tan^2 x)/(sec^3 x), but i dont know what to do next, because it doesnt easily simplfy...

2. integral (b = 1, a = 0) x(sqrt (x^2 + 4)) dx

3. integral (du)/(u(sqrt(5-u^2)))

4 integral (dt)/(sqrt(t^2 -6t +13))

for this one, i think we are supposed to complete the square and then evaluate the integral..

2. Originally Posted by skabani

1. integral (sqrt (x^2-a^2))/x^4 dx

i can get up to integral (a tan^2 x)/(sec^3 x), but i dont know what to do next, because it doesnt easily simplfy...
i got $\frac 1{a^2} \int \frac {\tan^2 \theta}{\sec^3 \theta}~d \theta$ (check this, i did it in my head )

if that's true, by using $\tan \theta = \frac {\sin \theta}{\cos \theta}$ and $\sec \theta = \frac 1{\cos \theta}$, this simplifies to $\frac 1{a^2} \int \sin^2 \theta \cos \theta ~d \theta$

now do a substitution. let $u = \sin \theta$ and the rest should be simple

2. integral (b = 1, a = 0) x(sqrt (x^2 + 4)) dx
substitution, $u = x^2 + 4$

3. integral (du)/(u(sqrt(5-u^2)))
you can do a trig substitution here, $u = \sqrt{5} \sin \theta$

4 integral (dt)/(sqrt(t^2 -6t +13))

for this one, i think we are supposed to complete the square and then evaluate the integral..
yes, that can work. this becomes $\int \frac {dt}{\sqrt{(t - 3)^2 + 4}}$

now do a trig substitution. $t - 3 = 2 \tan \theta$

3. Hello, skabani!

$1)\;\;\int \frac{\sqrt{x^2-a^2}}{x^4}\,dx$

Let: $x \:=\:a\sec\theta\quad\Rightarrow\quad dx \:=\:a\sec\theta\tan\theta\,d\theta$

Substitute: . $\int\frac{(a\tan\theta)\,(a\sec\theta\tan\theta\,d \theta)}{a^4\sec^4\!\theta} \;=\;\frac{1}{a^2}\int\frac{\tan^2\!\theta}{\sec^3 \!\theta}\,d\theta$

. . $= \;\frac{1}{a^2}\int \frac{\frac{\sin^2\!\theta}{\cos^2\!\theta}} {\frac{1}{\cos^3\!\theta}}\,d\theta \;=\;\frac{1}{a^2}\int\sin^2\!\theta\cos\theta\,d\ theta$

Let: $u = \sin\theta\quad\Rightarrow\quad du \:=\:\cos\theta\,d\theta$

Substitute: . $\frac{1}{a^2}\int u^2\,du \;=\;\frac{1}{3a^2}u^3 + C$

Back-substitute: . $\frac{1}{3a^2}\sin^3\!\theta + C$

Back-substitute:

We have: . $\sec\theta \:=\:\frac{x}{a} \:=\:\frac{hyp}{adj} \quad\Rightarrow\quad opp = \sqrt{x^2-a^2} \quad\Rightarrow\quad \sin\theta \:=\:\frac{\sqrt{x^2-a^2}}{x}$

And we have: . $\frac{1}{3a^2}\left(\frac{\sqrt{x^2-a^2}}{x}\right)^3 + C \;=\;\frac{1}{3a^2}\cdot\frac{(x^2-a^2)^{\frac{3}{2}}}{x^3} + C$

4. thanks so much!!

did you happen to get any of the other ones, im having trouble with them too..

thanks again!!

5. Originally Posted by Jhevon
i got $\frac 1{a^2} \int \frac {\tan^2 \theta}{\sec^3 \theta}~d \theta$ (check this, i did it in my head )

if that's true, by using $\tan \theta = \frac {\sin \theta}{\cos \theta}$ and $\sec \theta = \frac 1{\cos \theta}$, this simplifies to $\frac 1{a^2} \int \sin^2 \theta \cos \theta ~d \theta$

now do a substitution. let $u = \sin \theta$ and the rest should be simple

substitution, $u = x^2 + 4$

you can do a trig substitution here, $u = \sqrt{5} \sin \theta$

yes, that can work. this becomes $\int \frac {dt}{\sqrt{(t - 3)^2 + 4}}$

now do a trig substitution. $t - 3 = 2 \tan \theta$

thanks...so for 2, i got 1/8 - 1/4sqrt(5) + c
3. 1/5 ln u/sqrt(5) ln (sqrt(5) + u)/sqrt(5-^2)) + c
4. 1/3 arctan (x-3)/2 + c

are these right?