微積分ヘルプ: ∫√(16-x^2)dx の積分 - 三角関数の置換による積分

integral of 1/sqrt(16-x^2) dx

微積分ヘルプ: ∫ x^2/√(16-x^2 ) dx の積分 - 三角関数置換による積分 - 解決済み!

integral of 1/sqrt(x^2+16)

Integral of 6/sqrt(16-x^2)

Integral 42: int 1/[x^2 sqrt(16 x^2)} dx

Integrate | 1/sqrt(16-x^2) dx

Evaluate the following integrals: ∫√(16 x^2+25) d x

`int_(0)^(4)sqrt(16-x^(2))dx=`

Integration of Rational Functions By Completing The Square - Calculus

Calculus Help: Integral of (x^3 dx)/√(16-x^2 ) - Integration by substitution

Calculus Help: ∫ dx/(x√(16x^2-1)) - Integration by trigonometric substitution - Techniques

Integration of square root of 16 minus x square, int(sqrt(16-x^2)).

`int(dx)/(sqrt(16-x^(2)))`

Integration of 1/√16x²+25

Find the domain and Range of the function f(x)=root(16-x^2) l Relation and Function 11th

Calculus: Integrating sqrt(16-16x^2) from 0 to 1, using substitution x = Sinθ

三角関数の置換積分 sqrt(x^2 - 16)/x

Calculus Help: Integral of ∫ √(16x^2+25) dx - Integration by trigonometric substitution

Integral of 1/sqrt(16 - x^2) dx using trigonometric substitution