The integral ∫ √16-9x2 dx equals

Question

The integral ∫ √16-9x2 dx equals (A)  (x/2)√16-9x2+(8/3)sin-1((3x/4))+C  (B)  (3x/2)√16-9x2+ 16 sin-1 ((3x/4))+C  (C)  (π/2) sin-1 ( (3x/4)) + (9x/2) +C  (D) None of the above

Tardigrade Admin · Accepted Answer

Let I = ∫ √16-9x2 dx = ∫ √9((16/9) - x2) dx = 3 ∫ √((4/3))2 - x2 dx = 3[(x/2)× (√16-9x2/3)+(((4/3))2/2) sin-1((x/(4)3))] + C = 3[(x/2)× (√16-9x2/3) + (16/9×2) sin-1((3x/4))]+C = (x/2)√16-9x2 + (8/3) sin-1((3x/4))+ C

Q. The integral $\int 16 - 9 x^{2}$ $d x$ equals

$\frac{x}{2} 16 - 9 x^{2} + \frac{8}{3} s i n^{- 1} (\frac{3 x}{4}) + C$

$\frac{3 x}{2} 16 - 9 x^{2} + 16 s i n^{- 1} (\frac{3 x}{4}) + C$

$\frac{π}{2} s i n^{- 1} (\frac{3 x}{4}) + \frac{9 x}{2} + C$

$N o n e o f t h e ab o v e$

Q. The integral ∫16−9x2​ dx equals

2x​16−9x2​+38​sin−1(43x​)+C

23x​16−9x2​+16sin−1(43x​)+C

2π​sin−1(43x​)+29x​+C

Noneoftheabove

Let I=∫16−9x2​dx =∫9(916​−x2)​dx =3∫(34​)2−x2​dx =3[2x​×316−9x2​​+2(34​)2​sin−1(34​x​)]+C =3[2x​×316−9x2​​+9×216​sin−1(43x​)]+C =2x​16−9x2​+38​sin−1(43x​)+C

Q. The integral $\int 16 - 9 x^{2}$ $d x$ equals

$\frac{x}{2} 16 - 9 x^{2} + \frac{8}{3} s i n^{- 1} (\frac{3 x}{4}) + C$

$\frac{3 x}{2} 16 - 9 x^{2} + 16 s i n^{- 1} (\frac{3 x}{4}) + C$

$\frac{π}{2} s i n^{- 1} (\frac{3 x}{4}) + \frac{9 x}{2} + C$

$N o n e o f t h e ab o v e$