∫ (cos t) ^ 2 DT = what?

∫ (cos t) ^ 2 DT = what?

The identity cos2x = 2cos & # 178; X - 1
So cos & # 178; X = (1 + cos2x) / 2
∫ (cost)² dt
= ∫ (1 + cos2t)/2 dt
= (1/2)∫ dt + (1/2)(1/2)∫ cos2t d(2t)
= (1/2)t + (1/4)sin2t + C

Finding the reciprocal of [∫ cos (T ^ 2) DT]
The integral interval is [0, root x]

Let T ^ 2 = y
The integral interval of ∫ cos (T ^ 2) DT is [0, root x]
=The integral interval of ∫ (cosy / (2t)) dy is [0, x]
=∫ (1 / 2) (cosy / (y ^ (1 / 2))) dy integral interval is [0, x]
So the reciprocal of [cos (T ^ 2) DT]
=(1/2)(cosx/(x^(1/2))

∫ DT / (COS ^ 2 (T / 2)) please explain

Exchange yuan
x=t/2
t=2x
dt=2dx
Original formula = ∫ sec ^ 2 (x) 2DX
=2∫sec^2(x)dx
=2tanx+C
=2tan(t/2)+C
The coefficient upstairs is wrong

Prove the integrand expression f (T) DT = f [b (x)] B '(x) - f [a (x)] a' (x)
A (x), B (x) are differentiable functions, f (T) are continuous functions
Wrong: the formula on the left side of the equal sign should be derived

Let a (x), B (x) be the binary function of the lower limit a and the upper limit B of the integral, and let H (a (x), B (x)). Then the derivative formula of the composite function, DH / DX = (&; H / &; a) a '(x) + (&; H / &; b) B' (x) = (- f (a (x))) a '(x) + F (b (x)) B' (x) = f (b (x)) B '(x) - F

It is known that the definite integral of F (TX) DT of T from 0 to 1 = 1 / 2F (x) + 1, and the continuous function f (x)
Sorry, the integral number can't be typed

∫[0,1]f(tx)dt=(1/2)f(x)+1
f(0)/2=-1,f(0)=-2
[(1/2)f(x)]'=f(x)
f(x)'/f(x)=2
dlnf(x)=2
lnf(x)=2x+C0
f(x)=C1*e^(2x)
f(0)=C1=-2
f(x)=-2e^(2x)

∫ (subscript 0, superscript 1) f (XT) DT = f (x) + xsinx

It seems that the derivation first

The primitive function of definite integral sin square MX

Cos 2mx = 1-2sin square MX. So sin square MX = half of 1-cos 2mx, so the integral is 1 / 2x-1 / 2msin 2mx. Very simple, just simplify it

How to find the original function of xsinx?
How to find the indefinite integral of xsinx?

A:
Partial integration method ∫ UDV = UV - ∫ VDU
∫ xsinx dx
= - ∫ x d(cosx)
=-xcosx+∫ cosx dx
=-xcosx+sinx+C

Definite integral problem: the first derivative of F (x) is greater than 0, and the second derivative is greater than 0. Q: which of the following is the largest area? A. the integral of F (x) from a to B. B. (B-A) f (b)
The first derivative of F (x) is greater than 0 and the second derivative is greater than 0. Q: which of the following is the largest area? A. the integral of F (x) from a to B. (B-A) f (b) C.1 / 2 (B-A) [f (a) + F (b)]

The first derivative of F (x) is greater than 0, which means that the function increases monotonically in the interval [a, b], and f (b) is the maximum. F (a) is the minimum. The second derivative is also greater than 0. It means that the curve is concave. So there is (B-A) f (b) > [f (b) - f (a)] (B-A) / 2 > the integral of F (x) from a to B

Let f (x) have a second order continuous derivative on (0,1), if f (π) = 2, ∫ (0 to π) [f (x) + F "(x)] sinxdx = 5, find f (0)

∫(0→π)f''(x)sinxdx=∫(0→π)sinxd(f'(x))
=sinxf'(x)|(0→π)-∫(0→π)f'(x)cosxdx
=-∫(0→π)cosxd(f(x))
=-cosxf(x)|(0→π)-∫(0→π)f(x)sinxdx
So the left = - cosxf (x) | (0 → π) = f (π) + F (0)
So f (0) = 5-F (π) = 3