Mathematics

Evaluate $$\displaystyle\int_{0}^{a}\sqrt{a^{2}-x^{2}}dx$$


SOLUTION
$$I=\displaystyle\int_{0}^{a}\sqrt{a^{2}-x^{2}}dx$$
Let $$x=a\sin t$$
Then, $$dx=a\cos t dt$$                               [ Differentiating both sides ]
When $$x=0,$$      
$$0=a\sin t$$
$$t=\sin^{-1}(0)$$
$$\therefore$$ $$t=0$$
When, $$x=a,$$
$$a=a\sin t$$
$$1=\sin t$$
$$t=\sin^{-1}(1)$$
$$\therefore$$  $$t=\dfrac{\pi}{2}$$

$$I=\displaystyle\int_{0}^{a}\sqrt{a^{2}-x^{2}}dx$$

$$\Rightarrow$$  $$I=\displaystyle\int_0^{\dfrac{\pi}{2}}\sqrt{(a^2-a^2\sin^2t)}a\cos t dt$$

          $$=\displaystyle\int_0^{\dfrac{\pi}{2}}\sqrt{a^2(1-\sin^2t)}a\cos t dt$$

          $$=\displaystyle\int_0^{\dfrac{\pi}{2}}\sqrt{a^2\cos^2 t}a\cos t dt$$

          $$=\displaystyle\int_0^{\dfrac{\pi}{2}}a\cos t.a\cos t dt$$

          $$=\displaystyle\int_0^{\dfrac{\pi}{2}}a^2\cos^2 t dt$$

         $$=a^2\displaystyle\int_0^{\dfrac{\pi}{2}} \dfrac{1+\cos 2t}{2}dt$$

         $$=\dfrac{a^2}{2}\left[t+\dfrac{\sin 2t}{2}\right]^{\dfrac{\pi}{2}}_0$$

         $$=\dfrac{a^2}{2}\left(\dfrac{\pi}{2}-0\right)$$

         $$=\dfrac{a^2\pi}{4}$$

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Subjective Medium Published on 17th 09, 2020
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