AQA Core 4 4.09 2sin(x) + 3cos(x) を Rsin(x+alpha) の形式で表す
微積分ヘルプ: 積分 ∫cosx/(3+2sinx+sin^2x ) dx - 三角関数の置換による積分
LAQ Video 6 Evaluate integral of 2sinx+3cosx+4/3sinx+4cosx+5 dx
integrate 2sin(x) + 3cos(x) dx
integration of 3cosx+2/sinx+2cosx+3
`int(2sinx+3cosx)/(3sinx+4cosx)dx`
Trigonometric equations Solve (3cos(x)-2sin(x))/(2cos(x)-3sin(x) )=1/2 Where 0≤x≤360
Express √3 sin𝑥−cos𝑥 in the form 𝑘 sin(𝑥−𝑎)
Evaluate: `int(2sinx+3cosx)/(3sinx+4cosx)dx`
微積分ヘルプ: この三角関数の値を求めなさい: sin2x+√3 cosx=0
#3.28|| Indefinite Integrals|| Integration of (2+ 3 cosx)/sin²x
`int(dx)/((2sinx+3cosx)^2)`
Solving 2sin(x)=1-2cos(x)
`int(cosx+2sinxdx)/(3cosx+4sinx)`
If ` y=cos ^(-1) ((3cos x-2sin x )/( sqrt(13))),then (dy)/(dx)=`
JEE MAINS 2018 Evaluate `int(3cosx+2)/(sinx+2cosx+3)`dx
`int(dx)/(3+2sinx+cosx)` dxJ 3 + 2 sin x + cos x'
Evaluate : `int(3cosx+2)/(sinx+2cosx+3)` dx
`(cos x)/(sqrt(sin^(2)x-2sinx-3))`
Evaluate: `int1/((2sinx+3cosx)^2)\ dx`
