Problems from Integrals with changes of variable

Development:

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

• $dx=3\cdot \cos (t)\cdot dt$

$x=0\Rightarrow u=arcsin({\displaystyle \frac{0}{3}})=0$

$x=3\Rightarrow u=arcsin({\displaystyle \frac{3}{3}})={\displaystyle \frac{\pi }{2}}$

• ${\displaystyle {\int }_0^3\sqrt{9-x^2}dx={\int }_0^{\frac{\pi }{2}}3\cdot \cos (t)\cdot \sqrt{9-3^2\cdot {\sin }^2(t)}dt=9{\int }_0^{\frac{\pi }{2}}{\cos }^3(t)dt}$

• ${\displaystyle 9{\int }_0^{\frac{\pi }{2}}{\cos }^2(t)dt=9{\int }_0^{\frac{\pi }{2}}{\displaystyle \frac{1+\cos (2t)}{2}}dt=}$

$={\displaystyle 9{\int }_0^{\frac{\pi }{2}}{\displaystyle \frac{1}{2}}dt+{\int }_0^{\frac{\pi }{2}}{\displaystyle \frac{\cos (2t)}{2}}dt={\displaystyle \frac{9}{4}}\pi +0={\displaystyle \frac{9}{4}}\pi }$

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:

${\displaystyle {\int }_0^{\frac{\pi }{2}}\sqrt{9-x^2}dx={\displaystyle \frac{9}{4}}\pi }$

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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