Question

How? solve this please

Answer 1

See process below

Explanation:

Lets aply the change `x=4sinu` then `dx=4cosudu` and

`sqrt(16-16sin^2)=sqrt(16(1-sin^2u))=4cosu`

In our integral `int(16sin^2u·cancel4cancelcosudu)/(cancel4cancelcosu)=int16sin^2udu=16I`

Attack this last integral by parts

`u=sinu` and `dv=sinudu` from here `v=-cosu` and `du=cosudu`

`16I=-sinucosu+intcos^2udu=-sinucosu+int(1-sin^2u)du=-sinucosu+u-intsin^2udu`. Thus, finally

`17I=-sinucosu+u` transposing terms `I=u/17-(sinucosu)/17`

Un-doing the change `u=arcsin(x/4)`

`I=1/17arcsin(x/4)-1/17·x/4sqrt(1-x^2/4^2)+C`

Answer 2

`8(pi/6-sqrt3/4)~~0.72469`

Explanation:

`int_0^2" "x^2/sqrt(16-x^2)" "dx`

Let `x=4sin(u)` and `dx=4cos(u)" "du`

`int_0^2" "(16sin^2(u)*4cos(u))/sqrt(16-16sin^2(u))" "du`

Simplify,

`int_0^2" "(16sin^2(u)*4cos(u))/(sqrt16*sqrt(1-sin^2(u)))" "du`

Refine,

`int_0^2" "(16sin^2(u)*cancel(4)cos(u))/(cancel(sqrt16)*sqrt(1-sin^2(u)))" "du`

Hence,

`int_0^2" "(16sin^2(u)cos(u))/sqrt(1-sin^2(u))" "du`

Apply trigonometric functions,

`int_0^2" "(16sin^2(u)cos(u))/sqrt(cos^2(u))" "du`

Simplify,

`int_0^2" "(16sin^2(u)cancel(cos(u)))/cancel(cos(u))" "du`

Hence,

`int_0^2" "16sin^2(u)" "du`

Apply trigonometric function,

`int_0^2" "16*(1-cos(2u))/2" "du`

Simplify and bring out the constant,

`8int_0^2" "1-cos(2u)" "du`

Integrate,

`8[u-1/2sin(2u)]_0^2`

Substitute `u=arcsin(x/4)`

`8[arcsin(x/4)-1/2sin(2arcsin(x/4))]_0^2`

Compute boundaries,

`8[(arcsin(1/2)-1/2sin(2arcsin(1/2)))-arcsin(0)-1/2sin(2arcsin(0)))]`

Solve,

`8(pi/6-sqrt3/4)~~0.72469`