# Problems from Integrals with changes of variable

Calculate the following integral by the method of the change of variable: $$\displaystyle \int_0^3 \sqrt{9-x^2} \ dx$$

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### Development:

• Change of variable $$x=3 \cdot \sin(t)$$

• $$dx=3 \cdot \cos(t) \cdot dt$$

$$x=0 \Rightarrow u=arcsin\Big(\dfrac{0}{3}\Big)=0$$

$$x=3 \Rightarrow u=arcsin\Big(\dfrac{3}{3}\Big)=\dfrac{\pi}{2}$$

• $$\displaystyle \int_0^3 \sqrt{9-x^2} \ dx= \int_0^\frac{\pi}{2} 3\cdot\cos(t)\cdot\sqrt{9-3^2\cdot\sin^2(t)} \ dt =9\int_0^\frac{\pi}{2}\cos^3(t) \ dt$$

• $$\displaystyle 9 \int_0^\frac{\pi}{2} \cos^2(t) \ dt=9 \int_0^\frac{\pi}{2} \dfrac{1+\cos(2t)}{2} \ dt=$$

$$=\displaystyle 9\int_0^\frac{\pi}{2} \dfrac{1}{2} \ dt+\int_0^\frac{\pi}{2}\dfrac{\cos(2t)}{2} \ dt = \dfrac{9}{4}\pi+0=\dfrac{9}{4}\pi$$

Let's observe that, when we have an integral with a sine or a cosine raised to an even exponent, we will apply the formula of the double angle as many times as necessary in order to reduce the degree of the integral.

### Solution:

$$\displaystyle \int_0^\frac{\pi}{2} \sqrt{9-x^2} \ dx=\dfrac{9}{4}\pi$$

As we can see, if the integration limits are correctly done, it is not necessary to undo the change of variable.

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