Nilai [tex] \int\limits^5_0 {} \sqrt{25- x^{2} } \, dx [/tex] adalah ... A. [tex] \frac{25 \pi }{4} [/tex] B. [tex] \frac{25 \pi }{8} [/tex] C. [tex] \frac{25

lanjutan:
Gunakan teorema sin(arc sin(x/5)) = x/5 dan cos(arc sin(x/5))=akar(1-(x^2-25))
Sehingga diperoleh
integral akar (25-x^2) dx = (5x akar (1-(x^2/25)))/2 + (25 arc sin (x/5))/2
Substitusi batas atas integral yaitu 5 (batas bawah 0 menghasilkan 0 shg diabaikan)
shg diperoleh
= 0 + (25 arc sin 1)/2 = 0 + (25×pi/2)/2 = 25pi/4.
(Jawaban A)

[tex]\displaystyle \text{misal :}\\x=5\sin t\\dx=5\cos t\,dt\\\\\int\limits^5_0\sqrt{25-x^2}\,dx=\int\limits^5_0\sqrt{25-25\sin^2t}\cdot5\cos t\,dt\\\int\limits^5_0\sqrt{25-x^2}\,dx=\int\limits^5_05\cos t\cdot5\cos t\,dt\\\int\limits^5_0\sqrt{25-x^2}\,dx=25\int\limits^5_0\cos^2t\,dt\\\int\limits^5_0\sqrt{25-x^2}\,dx=25\int\limits^5_0\frac{\cos2t+1}{2}\,dt\\\int\limits^5_0\sqrt{25-x^2}\,dx=\left\frac{25}{4}\sin2t+\frac{25}2t\right|^5_0\\\int\limits^5_0\sqrt{25-x^2}\,dx=\left\frac{25\sin t\cos t}{2}+\frac{25}{2}t\right|^5_0[/tex]
[tex]\displaystyle \int\limits^5_0\sqrt{25-x^2}\,dx=\left\frac{5x\sqrt{1-\frac1{25}x^2}}{2}+\frac{25}{2}\arcsin\frac15x\right|^5_0\\\int\limits^5_0\sqrt{25-x^2}\,dx=\left\frac{x\sqrt{25-x^2}+25\arcsin\frac15x}{2}\right|^5_0\\\int\limits^5_0\sqrt{25-x^2}\,dx=\frac{5\sqrt{25-5^2}+25\arcsin\frac15(5)}{2}-\frac{0\sqrt{25-0^2}+25\arcsin\frac15(0)}{2}[/tex]
[tex]\displaystyle \int\limits^5_0\sqrt{25-x^2}\,dx=\frac{0+25\arcsin1}{2}-\frac{0+25\arcsin0}{2}\\\int\limits^5_0\sqrt{25-x^2}\,dx=\frac{25\cdot\frac12\pi}{2}-\frac{25\cdot(0)}{2}\\\int\limits^5_0\sqrt{25-x^2}\,dx=\frac{25\cdot\frac12\pi}{2}\\\boxed{\boxed{\int\limits^5_0\sqrt{25-x^2}\,dx=\frac{25\pi}{4}}}[/tex]