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

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!

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

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

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

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

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

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

Part 153 Integration of 1/√(16-6x-x²)dx, 1/√(7-6x-x²)dx, 1/√(5x²-2x)dx

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

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

Integral of 1/(16+x^2) || Integration by Trigonometric Substitution

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

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

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

三角関数の置換積分 sqrt(x^2 - 16)/x

Integration of 1/√16x²+25

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

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

integration of sqrt(16-x^2) and sqrt(x^2-16)