Calculation of definite integral (1 / 2 ~ 1) arcsinx ^ (1 / 2) / (x (1-x)) ^ 1 / 2DX

Calculation of definite integral (1 / 2 ~ 1) arcsinx ^ (1 / 2) / (x (1-x)) ^ 1 / 2DX

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For the indefinite integral of arcsinx / x ^ 2, I calculate = - arcsinx / x + ln | (1 / x) - √ (1-x & # 178;) / X | + C
But is the answer - 1 / 2 (xcsc ^ 2x + cscxcotx) + C wrong?

Wrong answer

Let's help you, find the indefinite integral ∫ coslnx =? (partial integral method)

First do the transformation LNX = t, x = e ^ t, DX = e ^ TDT,
∫coslnxdx=∫cost*e^tdt,
We'll do it two more times,
∫cost*e^tdt=e^t*sint-∫sint*e^tdt
=e^t*sint-[-e^t*cost+∫cost*e^tdt],
transposition,
2∫cost*e^tdt=e^t(sint+cost)+2C,
∫cost*e^tdt=e^t(sint+cost)/2+C,
∫coslnxdx=x(sinlnx+coslnx)/2+C.
Do not do the first step of transformation, direct integration can also be divided, but not as clear after transformation

In indefinite integral, there is a sequence of partial integral. What is it?

∫uv'dx= uv - ∫u'vdx

The following indefinite integrals 1) ∫ xsin2xdx 2) ∫ xlnxdx 3) ∫ arccosxdx 4) ∫ xarctanxdx can be obtained by the method of partial integration
The following indefinite integrals can be obtained by the method of partial integration
1）∫xsin2xdx
2）∫xlnxdx
3）∫arccosxdx
4）∫xarctanxdx

2) 3) the answer is the same as upstairs, 1) ∫ xsin2xdx = (- 1 / 2) ∫ xdcos2x = (- 1 / 2) xcos2x + (1 / 2) ∫ cos2xdx = (- 1 / 2) xcos2x + (1 / 4) sin2x + C2) ∫ xlnxdxdx = (1 / 2) ∫ lnxdx ^ 2 = (1 / 2) x ^ 2lnx - (1 / 2) ∫ xdx = (1 / 2) x ^ 2lnx - (1 / 4) x ^ 2 + C3) ∫ arccosxdx = xarccosx -

It is necessary to find the following indefinite integral ∫ x ^ 3 times e ^ x times DX by the method of partial integration

∫ X & # 179; e ^ x DX = ∫ X & # 179; de ^ x, the first time of integration by parts
=X & # 179; e ^ X - ∫ e ^ xdx & # 179; = x & # 179; e ^ X - 3 ∫ X & # 178; e ^ xdx, the first integration by parts
=X & # 179; e ^ X - 3 ∫ X & # 178; de ^ x, the second integration by parts
=X & # 179; e ^ X - 3x & # 178; e ^ x + 3 ∫ e ^ xdx & # 178; = x & # 179; e ^ X - 3x & # 178; e ^ x + 6 ∫ Xe ^ xdx, the second integration by parts
=X & # 179; e ^ X - 3x & # 178; e ^ x + 6 ∫ xde ^ x, the third integration by parts
=X & # 179; e ^ X - 3x & # 178; e ^ x + 6xe ^ X - 6 ∫ e ^ xdx, the third integration by parts
= x³e^x - 3x²e^x + 6xe^x - 6e^x + C
= (x³-3x²+6x-6)e^x + C

If x is an acute angle, sin2x = a, then SiNx + cosx=

We can see that 2sinxcosx = a
So the original formula = the square of SiNx + cosx under the root sign
=Square of SiNx + square of cosx + 2sinxcosx under root sign
Just put it in
=A + 1 under root

If | 2x-1 | = 2x-1, | 3x-5 | = 5-3x
|It's absolute value roar!

2x-1 | = 2x-1, 2x-1 is greater than or equal to 0, X is greater than or equal to 1 / 2
3x-5 | = 5-3x, 3x-5 is less than or equal to 0, and X is less than or equal to 5 / 3
So 1 / 2 less than or equal to x less than or equal to 5 / 3

Given that the root of the equation x + KX − 2 = 2 about X is less than 0, the value range of K is obtained

The denominator of the fractional equation is removed to get: x + k = 2X-4, the solution is: x = K + 4, according to the meaning of the question: K + 4 < 0, the solution is: K < - 4

In [0,2 π], the value range of X which makes SiNx > cosx hold is ()
A. (π4，π2)∪(π，5π4)B. (π4，π)C. (π4，5π4)D. (π4，π)∪(5π4，3π2)

In [0,2 π], draw the trigonometric function lines corresponding to SiNx and cosx, which are MT and OM, as shown in the figure: if SiNx > cosx is satisfied in [0,2 π], that is Mt > OM, so the range of X is: (π 4,5 π 4), so select C