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Trig substitution integration?

  • Thread starter emlekarc
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  • #1
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Homework Statement



Integrate dx/((x^2+1)^2)


Homework Equations



Tan^2=sec^2-1


The Attempt at a Solution



So I let x=tanx then dx=sec^2x


Then plugging everything in;

Sec^2(x)/(tan^2+1)^2

So it's sec^2/(sec^2x)^2) which is sec^2x/sec^4x

Canceling out the sec^2 gives you

1/sec^2x

Integrate: ln(sec^2x)+C

Is that right?
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement



Integrate dx/((x^2+1)^2)


Homework Equations



Tan^2=sec^2-1


The Attempt at a Solution



So I let x=tanx then dx=sec^2x
Don't use the same letter in a change of variables. Let ##y=\tan x## so ##dy=\sec^2x dx##.
[Edit, corrections follow] I meant ##x = \tan y## so ##dx =\sec^2 y dy##.
Then plugging everything in;

Sec^2(x)/(tan^2+1)^2

So it's sec^2/(sec^2x)^2) which is sec^2x/sec^4x

Canceling out the sec^2 gives you

1/sec^2x

Integrate: ln(sec^2x)+C

Is that right?
No. That isn't its antiderivative.
 
Last edited:
  • #3
benorin
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Homework Statement



Integrate dx/((x^2+1)^2)


Homework Equations



Tan^2=sec^2-1


The Attempt at a Solution



So I let x=tanx then dx=sec^2x


Then plugging everything in;

Sec^2(x)/(tan^2+1)^2

So it's sec^2/(sec^2x)^2) which is sec^2x/sec^4x

Canceling out the sec^2 gives you

1/sec^2x

Integrate: ln(sec^2x)+C

Is that right?
Like the man said, use another variable: Let $x=\tan\theta\Rightarrow dx=\sec^2\theta d\theta$ and the integral becomes

[tex]\int\frac{dx}{\left(1+x^2\right) ^2}=\int\frac{\sec^2\theta d\theta}{\left(1+\tan^2 x\right) ^2}=\int\frac{d\theta}{\sec^2\theta}=\int\cos^2\theta\, d\theta[/tex]

now use the identity [itex]\cos^2\theta=\frac{1}{2}\left( 1+\cos 2\theta\right)[/itex] to integrate w.r.t. [itex]\theta[/itex] then use [itex]x=\tan\theta[/itex] to rewrite the functions of [itex]\theta[/itex] as functions of [itex]x[/itex].
 

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