Evaluate ` int sqrt(1+sin. x/2 ) dx `

積分　(sinx)2乗分の1　の計算過程　Integral of 1/sin^2(x)

sqrt[1 + sin(x/2)] の不定積分を求めます。

Evaluate : integrate sqrt(1 - sin 2x) dx

`int sqrt(1 + sin 2x) dx` =

次を示します: ` int sqrt(1+ sin x ) dx = 2 (sin. x/2 - cos. x/2 )+ c = 2 sqrt( 1 - sin x ) + c `

`int1/(1-sinx/2)dx`

Integrate 0 to pi/2 sqrt(1-sin 2x) dx

`int sqrt (1 -sin 2x) dx` を積分します。

Integral of 1/(1+sin^2(x)) (substitution + substitution)

\[ \int \sqrt{1+\sin x} d x \] \( P \)

integrate sqrt(1 - sin x) dx

Integral of sqrt(1 - sin x) dx

Integral Practice #57: integral of (((cosx)^(2))/(1 - sinx))dx

Integral of 1/(1+sin(2x)), How to integrate, Indefinite Integral, Integration, Calculus

Integral Practice # 15: indefinite integral of ((1-sinx)/(1+sinx))dx (trigo identity and integral)

Q74 | Evaluate ∫ √(1+sin⁡x) dx | Integral of square root 1 + sin x | Integration under root 1+ sinx

Q75 | Evaluate ∫ √(1-sin⁡x) dx | Integral of square root 1 - sin x | Integration under root 1 - sinx

Find `int(xsin^(-1)x)/(sqrt(1-x^2))dx`...

integrate sqrt(1 + sin x) dx