Inverse Trig Substitution Integral

[Bazzinga]

\int x^{2} \sqrt{4+x^{2}} dx

I've already subbed in:

x = 2tan\theta
dx = 2sec^{2}\theta d\theta

and I've gotten down to:

16 \int tan^{3}\theta sec^{3}\theta d\theta

But now I have noo idea what to do! Can someone give me a hint?

 

[tiny-tim]

Hi Bazzinga!

(where did your tan³ come from? )

sorry, but you just have to slog it out, using repeated integration by parts

substituting x = sinh(u) instead would have been a lot easier

 

[rock.freak667]

I think your integral should reduce to
<br /> 16 \int tan^{2}\theta sec^{3}\theta d\theta <br />

as the x² term outside the square root will give tan².

Replace tan²θ with sec²θ -1 and try to simplify it. I think the integrals may get a bit complicated.

 

[SammyS]

...
and I've gotten down to:

16 \int \tan^{3}\theta \sec^{3}\theta d\theta

I think your integral should reduce to
<br /> 16 \int \tan^{2}\theta \sec^{3}\theta d\theta <br />

as the x² term outside the square root will give tan².

Replace tan²θ with sec²θ -1 and try to simplify it. I think the integrals may get a bit complicated.

Hi Bazzinga.

The integral:
16 \int \tan^{3}\theta \sec^{3}\theta \,d\theta
works out very easily with rock.freak667's suggestion for a replacement.

However, if rock.freak667 is right (probably the case) then your integral reduces to
<br /> 16 \int \sec^{5}\theta -\sec^{3}\theta d\theta <br />

 

[Bazzinga]

Oh sorry, the term is x^{3} not x^{2}

 

[SammyS]

Oh sorry, the term is x^{3} not x^{2}

In that case,
<br /> \int \tan^{3}\theta \sec^{3}\theta \,d\theta = \int \tan\theta\,\left(\sec^{5}\theta-\sec^{3}\theta \right)\, d\theta<br />

=\int \left(\sec^{4}\theta-\sec^{2}\theta \right)\tan\theta\,\sec\theta\ d\theta
​

This is quite straight forward.