Integration by trigonometric substitution

**db5vry** · September 13th 2009, 12:04 AM

I'm new to this type of question and I can only get so far into the question before I don't know what to do - I haven't been taught more but I very much would like to know.

It asks:

Use the substitution $x$ = $3\sin\theta$ to show that

$\int_{1.5}^{3} \sqrt{9 - x^2} dx$ = $\int_{a}^{b} k \cos^2\theta d\theta$ where the values of a, b and k are to be found.

Hence evaluate $\int_{1.5}^{3} \sqrt{9 - x^2} dx$ .

So far I can say that $x$ = $3\sin\theta$ and $dx$ = $3\cos\theta d\theta$

$\sqrt{9 - (3\sin\theta)^2}$ = $\sqrt{9 - 9\sin^2\theta}$

= $\sqrt{9 (1 - \sin^2\theta)}$
= $\sqrt{9\cos^2\theta}$
= $3\cos\theta$

And this is as far as I can get.

Could someone please help me finish the question and understand how it works from this point onwards?
Any help is greatly appreciated

**JML** · September 13th 2009, 03:14 AM

$\int_{1.5}^{3} \sqrt{9 - x^2} dx$ = $\int_{a}^{b} k \cos^2\theta d\theta$ where the values of a, b and k are to be found.

As you have done we let: $x$ = $3\sin\theta$ and $dx$ = $3\cos\theta d\theta$

so

$\int_{1.5}^{3} \sqrt{9 - x^2} dx$ = $\int_{1.5}^{3}\sqrt{9 - (3\sin\theta)^2} 3\cos\theta d\theta$

= $\int_{1.5}^{3} 3\cos \theta \cdot 3\cos\theta d\theta$

= $\int_{1.5}^{3} 9 \cos^2\theta d\theta$
so a=1.5 b=3 k=9

now to evaluate this integral you can use the identity: $\cos^2 x = \frac{1+cos(2x)}{2}$

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