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Find the indefinite integral of SiNx divided by (SiNx + cosx)
(SiNx) ^ 3 / (cosx + SiNx) DX = 1 / √ 2 * (SiNx) ^ 3 / (sin45 * cosx + cos45 * SiNx) DX = 1 / √ 2 * (SiNx) ^ 3 / sin (45 + x) DX set 45 + x = t = DX / dt = 1 = 1 / √ 2 * (sin (t-45)) ^ 3 / sin DX = 1 / √ 2 * ∫ (sin (t-45)) ^ 3 / sin t DX = 1 / √ 2 * ∫ (Sint / √ 2-cost / √ 2) ∫ (2) cost / √ 2) ∫ (Sint / √ 2-cost / √ 2) / (2) / (2) / (2) / (2) / (2) / (^ 3 / Sint DX will
Let's get the indefinite integral to the - x power of SiNx times E?
∫sinxe^(-x)dx
=-∫sinxde^(-x)
=-sinxe^(-x)+∫e^(-x)dsinx
=-sinxe^(-x)+∫cosxe^(-x)dx
=-sinxe^(-x)-∫cosxde^(-x)
=-sinxe^(-x)-cosxe^(-x)+∫e^(-x)dcosx
=-sinxe^(-x)-cosxe^(-x)-∫e^(-x)sinxdx
therefore
The original formula = - 1 / 2sinxe ^ (- x) - 1 / 2cosx * e ^ (- x) + C
=-1/2e^(-x)(sinx+cosx)+c
Find the square of the indefinite integral SiNx times the fourth power of cosx
This is not very difficult, (SiNx) ^ 2 * (cosx) ^ 4 = 1 / 4 (sin2x) ^ 2 (1 + cos2x) / 2 = (1 / 16) (1 + cos2x) (1-cos4x)
Then expand it, draw cos 2x cos4x with the sum and difference formula of product. Finally, it comes out
SiNx quartic + cosx quartic = 1-2sinx quadratic, cosx quadratic
SiNx quartic + cosx quartic
=The fourth power of SiNx + 2Sin? Xcos? X + the fourth power of cosx - 2Sin? Xcos? X
=(sin²x+cos²x)²-2sin²xcos²x
=1-2sin²xcos²x
ν SiNx quartic + cosx quartic = 1-2sinx quadratic cosx quadratic
F (x) = sin2 power (x + pai-4) - sin2 power (x-pai-4) minimum positive period
f(x)=sin²(x+π/4)-sin²(x-π/4)
=[1-cos(2x+π/2)]/2-[1-cos(2x-π/2)]/2
=[1+sin(2x)]/2-[1-sin(2x)]/2
=sin2x
The minimum positive period = 2 π / 2 = π
This is my conclusion after meditation,
If you can't ask, I will try my best to help you solve it~
If you are dissatisfied and willing, please understand~
Lim x tends to 0 and has (SiNx / x) to the (1 / (x ^ 2)) power. Find the limit Why is the answer to the (- 1 / 6) power of E?
When x tends to 0, SiNx = X-1 / 6x ^ 3 + O (x)
lim（1-1/6x^2）(1/x^2)=e(-1/6)
How to calculate the TaNx power of LIM (x tends to π / 2) (SiNx)?
One of the solutions to this problem is: if you want to know LIM (x - > π / 2) [(sinx-1) TaNx] = LIM (x - > π / 2) {[(sinx-1) / cosx] SiNx] SiNx} = LIM (x - > π / 2) [(sinx-1) / cosx] * LIM (x - > π / 2) (SiNx) = LIM (x - > π / 2) (SiNx = > π / 2) {[sin (x / 2) - cos (x / 2)] / [cos (x / 2) + sin (x / 2)]} * 1 = 0 * 1 = 0 * 1 = 0 LIM (x - / (x - [x - [x / 2) + SiNx (x / 2) + SiNx (x / 2)] 1 = 0> π / 2) {(SiNx) ^ [
LIM (1 + TaNx) to the power of 3 / SiNx, when x tends to o, why is the limit to the third power of E?
3 / SiNx power of LIM (1 + TaNx) = 1 / TaNx * 3tanx / SiNx power of LIM (1 + TaNx)
=The 3tanx / SiNx power of LIM (x - > 0) [(1 + TaNx) to the power of 1 / TaNx]
=The power of LIM (x - > 0) 3tanx / SiNx of e
=The power of LIM (x - > 0) 3x / X of e
=The third power of e
Using equivalent infinitesimal to find the limit LIM (TaNx SiNx) / SiNx ^ 3
The original formula = TaNx (1-cosx) / SiNx ^ 3 when x --- 0, then 1-cosx --- (1 / 2) x ^ 2 TaNx --- x SiNx ^ 3 --- x ^ 3 = x * (1 / 2) x ^ 2 / x ^ 3 = 1 / 2 is expected to be adopted
Using Equivalent Infinitesimal Substitution to find Lim [(TaNx SiNx) / sin? 3x] limit when x tends to 0
So, Lim [(TaNx SiNx) / sin? 2x] = Lim [(SiNx / cosx SiNx) / 9x ^ 2] = Lim [SiNx (1-cosx) / 9x ^ 2] = Lim [SiNx (1-cosx) / 9x ^ 2] = Lim [SiNx (1-cosx) / 9x ^ 2 cosx]
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