求（sinx）的6次方的积分两种降幂方法：第一种：sin⁶x。= (sin²x)³。= [(1 - cos2x)/2]³。= (1/8)(1 - 3cos2x + 3cos²2x - cos³2x)。= 1/8 - (3/8)cos2x + (3/8

sinx=\frac{e^{ix}-e^{-ix}}{2i} ∴sin^{2n}x=\frac{(e^{ix}-e^{-ix})^{2n}}{(-4)^n} 考虑 (e^{ix}-e^{-ix})^{2n}=\sum_{k=0}^{2n}{C_{2n}^{k}e^{ix(2n-k)}(-1)^ke^{-kix}} =\sum_{k=0}^{2n}{C_{2n}^{k}(-1)^ke^{2ix(n-k)}}=\sum_{k=0}^{n-1}{C_{2n}^{k}(-1)^ke^{2ix(n-k)}}+(-1)^kC_{2n}^{n}+\sum_{k=n+1}^{2n}{C_{2n}^{k}(-1)^ke^{2ix(n-k)}}

\csc (-\frac{53\pi }{6}) prove\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x) prove\:\cot(2x)=\frac{1-\tan^2(x)}{2\tan(x)} prove\:\csc(2x)=\frac{\sec(x)}{2\sin(x)} Show More

sint的六次方的不定积分怎么求?∫ (sint)^6 dx=∫ [(sint)^2]^3 dx=(1/8)∫ [ 1- cos2t ]^3 dx= (1/8)∫ [ 1- 3cos2t + 3(cos2t)^2 - (cos2t)^3 ]dx=(1/8)[ t - (3/2)sin2t] +(3/8)∫ (cos2t)^2 百度首页

Mar 9, 2018 · sin^6(x)=-1/32(cos(6x)-6cos(4x)+15cos(2x)-10) We want to expand sin^6(x) One way is to use these identities repeatedly sin^2(x)=1/2(1-cos(2x)) cos^2(x)=1/2(1+cos(2x)) This often gets quite long, which sometimes leads to mistake Another way is to use the complex numbers (and Euler's formula) We can express sine and cosine as color(red)(sin(x)=(e ...

免费的数学问题解答者采用逐步解题讲解方式回答您的代数、几何、三角集合学、微积分以及统计学家庭作业问题，就像一位数学辅导老师那样为您提供帮助。

对此函数求积分 (sin(x))^6 自变量为 x: ((3*(sin(4*x)/2+2*x))/16-(sin(2*x)-sin(2*x)^3/3)/8-(3*sin(2*x))/8+x/4)/2+C 绘制该函数的图像 编辑该公式 本页链接

The calculator will find the integral/antiderivative of sin(x)^6, with steps shown.

Dec 17, 2023 · (sinx)^6的不定积分怎么求？答：(sinx)^6=[(sinx)^2]^3=[(1-cos2x)/2]^3=1/8(1-3cos2x+3(cos2x)^2-(cos2x)^3)=1/8(1-3cos2x+3(1+cos4x)/2-cos2x(1-(sin2x)^2))所以不定积 …

SHORT ANSWER: Yes, you can use cases, but you should use three cases. The first case is \sin x=0, the second is \cos x=0 (since that is also a denominator in your equation), and the third is ...