Question

Integral of sqrt(81-x^2)dx using the substitution of x=9sint

Answer 1

`x=9sin(t)\\dx=9cos(t)\,dt\\ \int\limits {√((81-x^2))} \, dx =\int\limits {√((81-(9sin(t))^2)} \, 9cos(t)\,dt\\ =\int\limits {\sqrt{(81-81\sin^2{t})}} \, 9cos(t)\,dt =\int\limits {\sqrt{81(1-\sin^2{t})}} \, 9cos(t)\,dt\\=\int\limits {(\sqrt{81\cos^2{t}})} \, 9cos(t)\,dt =\int\limits {(9cos(t))} \, 9cos(t)\,dt =\int\limits {81\cos^2{t}}\,dt\\ =(81)/(2)\int\limits {(1-cos(2t)})\,dt =(81)/(2)(t-(sin(2t))/(2)})+c\\ =(81)/(2)t-(81sin(2t))/(4)}+c`