1. ## Integration by substitution

integrate&#91;1&#47;&#40;x&#94;2&#40;x&#94;2 &#43;1&#41;&#94;0.5&#41;&#93; - Wolfram|Alpha

I don't entirely understand the first step. Could someone elaborate? I understand the basic idea of integration by sunstitution.

2. ## Re: Integration by substitution

Originally Posted by Stuck Man
integrate&#91;1&#47;&#40;x&#94;2&#40;x&#94;2 &#43;1&#41;&#94;0.5&#41;&#93; - Wolfram|Alpha

I don't entirely understand the first step. Could someone elaborate? I understand the basic idea of integration by sunstitution.
See the second bullet point: Trigonometric substitution - Wikipedia, the free encyclopedia

Since your integrand has 1^2 + x^2 in it, tangent is the appropriate trig substitution.

3. ## Re: Integration by substitution

Or

$\displaystyle \int \frac{dx}{x^2\sqrt{x^2+1}}= \int\frac{dx}{x^3\sqrt{1+ x^{-2}}}= \int (1+x^{-2})^{-1/2}x^{-3}\,dx$

Now make this substitution $\displaystyle 1+x^{-2}=u~\Rightarrow~x^{-3}dx=-(1/2)du$.

4. ## Re: Integration by substitution

I was originally trying the substitution u=tan x. I don't understand how the expression cot(u) csc(u) is obtained.

5. ## Re: Integration by substitution

Originally Posted by Stuck Man
I was originally trying the substitution u=tan x. I don't understand how the expression cot(u) csc(u) is obtained.
Let x = tan(u),  then dx = sec^2(u) du .

So you have: $\displaystyle \int \frac{dx}{x^2\sqrt{x^2+1}}= \int\frac{\sec^2(u) du}{\tan^2(u)\sqrt{\tan^2(u)+1}}$

Simplify it from there.

6. ## Re: Integration by substitution

I cannot simplify 1/(sec u - cos u) to cot u csc u.

7. ## Re: Integration by substitution

Multiply top and bottom by cos u.

Then you have
(cos u)/(1 - cos^2 u)

Here, the denominator is sin^2 u.

But what is (cos u)/(sin u sinu)... ?

8. ## Re: Integration by substitution

Thanks. Now I cannot understand the last part, substituting back u= inv tan x.

9. ## Re: Integration by substitution

Wolfram mislead me there. I used the values of sec u and tan u to get csc u.

10. ## Re: Integration by substitution

Originally Posted by Stuck Man
integrate&#91;1&#47;&#40;x&#94;2&#40;x&#94;2 &#43;1&#41;&#94;0.5&#41;&#93; - Wolfram|Alpha

I don't entirely understand the first step. Could someone elaborate? I understand the basic idea of integration by sunstitution.
Personally, I nearly always prefer to simplify integrands to sines and cosines, because I find it easier to manipulate them to forms where I can use substitutions.

Make the substitution $\displaystyle \displaystyle x = \tan{\theta} \implies dx = \sec^2{\theta}\,d\theta$ and the integral becomes

\displaystyle \displaystyle \begin{align*} \int{\frac{1}{x^2\sqrt{1 + x^2}}\,dx} &= \int{\frac{1}{\tan^2{\theta}\sqrt{1 + \tan^2{\theta}}}\,\sec^2{\theta}\,d\theta} \\ &= \int{\frac{ \sec^2{\theta} }{ \tan^2{\theta} \sqrt{ \sec^2{\theta} } }\,d\theta} \\ &= \int{ \frac{ \sec^2{\theta} }{ \tan^2{\theta} \sec{\theta} } \,d\theta} \\ &= \int{ \frac{ \sec{\theta} }{ \tan^2{\theta} } \,d\theta } \\ &= \int{ \frac{ \frac{1}{ \cos{\theta} } }{ \frac{ \sin^2{\theta} }{ \cos^2{\theta} } } \,d\theta } \\ &= \int{ \frac{ \cos^2{\theta} }{ \sin^2{\theta} \cos{\theta} } \,d\theta} \\ &= \int{ \frac{ \cos{\theta} }{ \sin^2{\theta} } \,d\theta} \\ &= \int{\frac{1}{u^2} \,du} \textrm{ after making the substitution }u = \sin{\theta} \implies du = \cos{\theta}\,d\theta \\ &= \int{u^{-2}\,du} \\ &= \frac{u^{-1}}{-1} + C \\ &= -\frac{1}{u} + C \\ &= -\frac{1}{\sin{\theta}} + C \\ &= -\frac{1}{ \frac{ \tan{\theta} }{ \sqrt{ 1 + \tan^2{\theta} } } } + C \end{align*}

\displaystyle \begin{align*} &= -\frac{ \sqrt{ 1 + \tan^2{\theta} } }{ \tan{\theta} } + C \\ &= - \frac{ \sqrt{ 1 + x^2 } }{ x } + C\end{align*}

11. ## Re: Integration by substitution

How did you replace sin theta in the third last line? Is that a known identity?

12. ## Re: Integration by substitution

$\displaystyle 1+\tan^2\theta = \sec^2\theta$

13. ## Re: Integration by substitution

That doesn't appear to have anything to do with it.

14. ## Re: Integration by substitution

Originally Posted by Stuck Man
How did you replace sin theta in the third last line? Is that a known identity?
Yes it is a known identity...

Originally Posted by jgv115
$\displaystyle 1+\tan^2\theta = \sec^2\theta$
Using the hint jgv gave you...

\displaystyle \displaystyle \begin{align*} 1 + \tan^2{\theta} &\equiv \frac{1}{\cos^2{\theta}} \\ \frac{1}{1 + \tan^2{\theta}} &\equiv \cos^2{\theta} \\ \frac{1}{1 + \tan^2{\theta}} &\equiv 1 - \sin^2{\theta} \\ \sin^2{\theta} &\equiv 1 - \frac{1}{1 + \tan^2{\theta}} \\ \sin^2{\theta} &\equiv \frac{1 + \tan^2{\theta}}{1 + \tan^2{\theta}} - \frac{1}{1 + \tan^2{\theta}} \\ \sin^2{\theta} &\equiv \frac{\tan^2{\theta}}{1 + \tan^2{\theta}} \\ \sin{\theta} &\equiv \pm \frac{\tan{\theta}}{ \sqrt{ 1 + \tan^2{\theta} } } \end{align*}

15. ## Re: Integration by substitution

Thanks. Why does the + or - symbol get thrown away?

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