A triangle can be transformed into a parallelogram by cutting and complementing. How can we deduce the triangle area calculation formula

A triangle can be transformed into a parallelogram by cutting and complementing. How can we deduce the triangle area calculation formula

It is the area of the parallelogram after cutting and complementing, bottom * height. At this time, the bottom is exactly half of the largest side of the original triangle

Why do vectors follow the rule of parallelogram? Or are they obtained by experiment?

The experiment also has the parallelogram rule with the vector in mathematics

The angle between two sides of a triangle is known
ex
BD(->)=(cosn)i(->)+sin(n)j(->)+3k(->)
BC(->)=(cosn)i(->)-sin(n)j(->)-3k(->)
Finding the angle B between BD and BC
CoSb = BD (- >) * (dot product) BC (- >) / | BD | BC |;

Is such, the space vector seeks the included angle. Looked on the high number book to be possible

It is known that an original function of F (x) is SiNx / X. find ∫ XF '(x) DX

f(x)=(sinx/x)'=(cosx*x-sinx)/x²
∫xf'(x)dx=xf(x)-∫x'f(x)dx
=xf(x)-∫f(x)dx
=xf(x)-sinx/x
=cosx-2sinx/x

∫(1+sinx)/(1+cosx+sinx)dx

I'm sorry I didn't learn well
I can't understand the question

Integral of ∫ SiNx √ (1 + cosx ^ 2) DX

∫ SiNx √ (1 + cosx ^ 2) DX = - ∫ √ (1 + cosx ^ 2) dcosx uses y = cosx, there are = - ∫ √ (1 + y ^ 2) dy = - Y / 2 * √ (1 + y ^ 2) - 1 / 2 * ln (Y + √ (1 + y ^ 2)) + C and y = cosx, which is obtained by substitution; = - cosx / 2 * √ (1 + cosx ^ 2) - 1 / 2 * ln (cosx + √ (1 + cosx ^ 2)) + C

Integral ∫ DX / (SiNx * cosx)

∫dx / (sinx * cosx)
=1/2∫dx /sin2x
=1/4∫d(2x)/sin2x
=1/4ln(sec2x-ctan2x)+c
=1/4ln(tanx)+c

Integral sign SiNx / (SiNx + cosx) DX =?
The answer is x / 2-1 / 2 * (LN (SiNx + cosx)) + C

Let Tan (x / 2) = t, then SiN x = 2T / (1 + T ^ 2) cos x = (1-T ^ 2) / (1 + T ^ 2) DX = 2 / (1 + T ^ 2) DT ∫ SiNx / (SiNx + cosx) DX = ∫ 2T / (1 + T ^ 2) * 2 / (1 + T ^ 2) / [2T / (1 + T ^ 2) + (1-T ^ 2) / (1 + T ^ 2)] DT = - 4 ∫ T / [(1 + T ^ 2) (T + √ 2-1) (T - √ 2-1) DT

Seeking indefinite integral: xsinx / cos * 3x
Process and skills of this kind of questions
It's cubic. I don't know how to get the first step on the first floor and the second part on the second floor

Let me tell you
On the second floor
∫ xsinx / (cosx) ^ 3DX --- he missed DX
Where sinxdx = - dcosx
Just take it in
I don't know how to ask

Indefinite integral of sine function to the sixth power
Such as the title`
I need the problem-solving process. Thank you```

∫(sinx)^6dx
=(1/8)∫(1-cos2x)^3dx
=(1/8)∫[1-(cos2x)^3+3(cos2x)^2-3cos2x]dx
----------------------------------------
among
∫(cos2x)^3dx=(1/2)∫1-(sin2x)^2dsin2x=(1/2)[sin2x-(1/3)(sin2x)^3]
∫(cos2x)^2dx=(1/2)∫[cos4x+1]dx=(1/2)[(1/4)sin4x+x]
∫cos2xdx=(1/2)sin2x
----------------------------------------
=(1/8){x-(1/2)[sin2x-(1/3)(sin2x)^3]+(3/2)[(1/4)sin4x+x]-(3/2)sin2x}
=5x/2+(sin2x)^3/6+3sin4x/8-5sin2x/2