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
$$16 \int tan^{2}\theta sec^{3}\theta d\theta$$

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
$$16 \int \tan^{2}\theta \sec^{3}\theta d\theta$$

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
$$16 \int \sec^{5}\theta -\sec^{3}\theta d\theta$$

 

[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,
$$\int \tan^{3}\theta \sec^{3}\theta \,d\theta = \int \tan\theta\,\left(\sec^{5}\theta-\sec^{3}\theta \right)\, d\theta$$

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

This is quite straight forward.