Question

Evaluate the integral

1
2
sin ¹(x/4)) ² dx
| (si
- 1
2
[(sin ~'(x/4)) ² dx
√16-x²
2
√16-x²
11

Answer 1

Substitute
`y=\sin^(-1)\left(\frac x4\right)`, so that

`dy = (dx)/(4√(1-\left(\frac x4\right)^2)) = (dx)/(√(16 - x^2))`

Then the integral is

`\displaystyle \int (\left(\sin^(-1)\left(\frac x4\right)\right)^2)/(√(16-x^2)) \,dx = \int y^2 \, dy`

and by the power rule,

`\displaystyle \int y^2 \, dy = \frac13 y^3 + C`

so that the original integral is

`\displaystyle \int (\left(\sin^(-1)\left(\frac x4\right)\right)^2)/(√(16-x^2)) \,dx = \boxed{\frac13 \left(\sin^(-1)\left(\frac x4\right)\right)^3 + C}`