# keep getting slightly wrong answer for integration question

• Jun 3rd 2009, 11:46 AM
superdude
keep getting slightly wrong answer for integration question
"Use the substitution x=secθ where 0<=θ<π/2 to evaluate
(integrand) √(x^2-1)/x^4 dx."

so after substitution:
⌠ 2
⎮ √(SEC(θ) - 1)
⎮ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ dx (it still should be dx right? not dθ?)
⎮ 4
⌡ SEC(θ)

after using trig identities I get (sigma)(1-sin^2θ)sinθcosθdx
the I let u = sinθ and -du=cosθdx

so
⌠ 3
- ⌡ (u - u ) du
and that is -(u^2)/2+(u^4)/4 +C
I suspect this is where I mess things up.

and then I plug back in the value of u to get
-(sin^2θ)/2+(sin^4θ)/4+c

my CAS gives me SIN(θ)^2·(SIN(θ)^2 - 2)/4 but I don't think that's the same(Doh)
• Jun 3rd 2009, 11:56 AM
Jester
Quote:

Originally Posted by superdude
"Use the substitution x=secθ where 0<=θ<π/2 to evaluate
(integrand) √(x^2-1)/x^4 dx."

so after substitution:
⌠ 2
⎮ √(SEC(θ) - 1)
⎮ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ dx (it still should be dx right? not dθ?) (**) you need to switch the dx over
⎮ 4
⌡ SEC(θ)

after using trig identities I get (sigma)(1-sin^2θ)sinθcosθdx
the I let u = sinθ and -du=cosθdx

so
⌠ 3
- ⌡ (u - u ) du
and that is -(u^2)/2+(u^4)/4 +C
I suspect this is where I mess things up.

and then I plug back in the value of u to get
-(sin^2θ)/2+(sin^4θ)/4+c

my CAS gives me SIN(θ)^2·(SIN(θ)^2 - 2)/4 but I don't think that's the same(Doh)

With your substitution $x = \sec \theta$then $\int \frac{\sqrt{x^2-1}}{x^4}\, dx = \int \frac{\sqrt{\sec ^2 \theta -1}}{\sec ^4 \theta }\, \sec \theta \tan \theta d\theta = \int \frac{ \tan^2 \theta}{\sec ^3 \theta}\, d \theta = \int \sin^2 \theta \cos \theta\, d \theta$. You should be able to get the rest.
• Jun 3rd 2009, 11:57 AM
Chris L T521
Quote:

Originally Posted by superdude
"Use the substitution x=secθ where 0<=θ<π/2 to evaluate
(integrand) √(x^2-1)/x^4 dx."

so after substitution:
⌠ 2
⎮ √(SEC(θ) - 1)
⎮ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ dx (it still should be dx right? not dθ?)
⎮ 4
⌡ SEC(θ)

after using trig identities I get (sigma)(1-sin^2θ)sinθcosθdx
the I let u = sinθ and -du=cosθdx

so
⌠ 3
- ⌡ (u - u ) du
and that is -(u^2)/2+(u^4)/4 +C
I suspect this is where I mess things up.

and then I plug back in the value of u to get
-(sin^2θ)/2+(sin^4θ)/4+c

my CAS gives me SIN(θ)^2·(SIN(θ)^2 - 2)/4 but I don't think that's the same(Doh)

When you make the substitution $x=\sec\theta$, it follows that $\,dx=\sec\theta\tan\theta\,d\theta$.

Thus, the integral becomes $\int\frac{\sqrt{\sec^2\theta-1}\sec\theta\tan\theta\,d\theta}{\sec^4\theta}=\in t\frac{\tan^2\theta}{\sec^3\theta}\,d\theta=\int\s in^2\theta\cos\theta\,d\theta$

Now let $u=\sin\theta\implies\,du=\cos\theta\,d\theta$

The integral now becomes $\int u^2\,du=\tfrac{1}{3}u^3+C$

Back-substituting, we get $\tfrac{1}{3}\sin^3\theta+C$

Now since $x=\sec\theta$ and $0\leqslant\theta\leqslant\frac{\pi}{2}$, it follows that $\sin\theta=\frac{\sqrt{x^2-1}}{x}$ (verify).

Therefore, $\int\frac{\sqrt{x^2-1}}{x^4}\,dx=\frac{\left(x^2-1\right)\sqrt{x^2-1}}{3x^3}+C$.

Does this make sense?
• Jun 3rd 2009, 12:06 PM
Amer
Quote:

Originally Posted by superdude
"Use the substitution x=secθ where 0<=θ<π/2 to evaluate
(integrand) √(x^2-1)/x^4 dx."

$\int\frac{\sqrt{x^2-1}}{x^4}dx$

let $x=sec\theta$
$dx=sec\theta(tan\theta)d\theta$

$\int\left(\frac{\sqrt{sec^2\theta-1}}{sec^4(\theta)}\right)sec(\theta)tan(\theta)d\t heta$

$\int\left(\frac{\sqrt{tan^2(\theta)}}{sec^3(\theta )}\right)tan(\theta)d\theta$

$\int\left(\frac{tan^2(\theta)}{sec^3(\theta)}\righ t)d\theta$

$\int\left(\frac{sec^2(\theta)-1}{sec^3(\theta)}\right)d\theta$

$\int\left(\frac{sec^2(\theta)}{sec^3(\theta)}\righ t)d\theta-\int\frac{1}{sec^3(\theta)}d\theta$

I leave the rest for you ...best wishes

I am always late I start to answer then I see a lot of people post faster than me that's wonderful it is the best forum the questions solved sooooooo quickly I love this thanks to all of you (Rofl)(Rofl)(Rofl)
• Jun 3rd 2009, 12:59 PM
Krizalid
put $x=\frac1u$ and you'll get an easy integral to solve.
• Jun 3rd 2009, 01:03 PM
Jester
Quote:

Originally Posted by Krizalid
put $x=\frac1u$ and you'll get an easy integral to solve.

I was waiting for you to come up with one of your slick substitutions (Rofl)
• Jun 4th 2009, 12:21 PM
superdude
one thing I don't get
Quote:

Originally Posted by Chris L T521
Now since $x=\sec\theta$ and $0\leqslant\theta\leqslant\frac{\pi}{2}$, it follows that $\sin\theta=\frac{\sqrt{x^2-1}}{x}$ (verify).

Therefore, $\int\frac{\sqrt{x^2-1}}{x^4}\,dx=\frac{\left(x^2-1\right)\sqrt{x^2-1}}{3x^3}+C$.

Does this make sense?

I follow up to the point where I get (sin^3θ/3)+C .I don't understand where you get the $x=\sec\theta$ from, is that an identity? when I plug that in for the value of sinθ I still get a differnt answer
i.e. (√(x^2-1))^3/3+C ≠ $\int\frac{\sqrt{x^2-1}}{x^4}\,dx=\frac{\left(x^2-1\right)\sqrt{x^2-1}}{3x^3}+C$
• Jun 4th 2009, 07:18 PM
Amer
from this you have $sin\theta=\frac{\sqrt{x^2-1}}{x}$ right

then from the picture below you can get sec(theta)

Attachment 11760
• Jun 6th 2009, 07:26 PM
superdude
thanks. i get it