How to find the indefinite integral of COS ^ 3x? There are 2 Li's book p97 example 342 above the integral formula how to get? Think 2 Li's book some places omitted a lot. I'm not good, want to think for a long time. There is also the book to find water pressure problem, I think there is a problem ~ not multiplied by G

How to find the indefinite integral of COS ^ 3x? There are 2 Li's book p97 example 342 above the integral formula how to get? Think 2 Li's book some places omitted a lot. I'm not good, want to think for a long time. There is also the book to find water pressure problem, I think there is a problem ~ not multiplied by G

The original formula = f (cosx) square ● cosxdx = f (cosx) square dsinx = f (1-sinx Square) dsinx = fdsinx fsinx square dsinx = cosx-1 / 3 (the third power of SiNx) + C
Please accept

Finding indefinite integral (1) ∫ xdx / 1 + √ x (2) ∫ (√ 2 + x-x ^ 2) DX (3) ∫ DX / 1 + √ 1-x ^ 2
The answer in the book is (1) 2x √ X / 3-x + 2 √ x-2ln (1 + √ x) + C
(2)(9/8)arcsin(2x-1/3)+(2x-1/4)(√2+x-x^2)+c
(3)arcsinx-(x/1+√1-x^2)+c
The root of question (2) goes to the last except DX
Question (3) has been answered
I still don't quite understand question (2), because I'm a beginner and I've just come into contact with advanced mathematics. Please

(1) Look at the molecule of the integral formula, x = (x + x ^ (1 / 2)) - (x ^ (1 / 2) + 1) + (1 + x ^ (- 1 / 2)) - x ^ (- 1 / 2)
therefore
∫xdx/1+√x＝∫x^(1/2)dx+∫1dx+∫x^(-1/2)dx+∫[x^(-1/2)/(1+√x)]dx
∫xdx/1+√x＝(2/3)x^(3/2)+x+2x^(1/2)+2*ln(1+x^(1/2))
(2) What's wrong with this problem? What's the difficulty?
Isn't it equal to √ 2 * x + (1 / 2) * x ^ 2 - (1 / 3) * x ^ 3?
(3) . let x = Sint
∫dx/1+√1-x^2=∫cost/(1+cost)dt=t-∫1/(1+cost)dt
=t-∫[(1-cost)/(sint)^2]dt
=t+ctan(t)-1/sint
Then change t back to x ~!
arcsinx+[√1-(x^2)]/x-1/x
I made a mistake in the sign of the last item in the first question, but there was no big mistake
I forgot to add "C" to the other 1 and 3
I don't understand the second question. I don't know where your root sign ends. It's not clear. If you have any questions, leave a message in hi~
1、 If you don't understand the three questions, you can discuss them in hi
The second problem is not difficult, change the element, let y = x - 1 / 2, then use partial integral, and finally change back
You try first, no, leave a message
The second question is as follows:
Let y = X-1 / 2, then the integral is transformed into, ∫ (√ 9 / 4-y ^ 2) dy
Then integral by parts to get y (√ 9 / 4-y ^ 2) - ∫ y (√ 9 / 4-y ^ 2)'dy
The above formula is y (√ 9 / 4-y ^ 2) + ∫ [(y ^ 2) / (√ 9 / 4-y ^ 2)] dy
Continue: Y (√ 9 / 4-y ^ 2) - ∫ (√ 9 / 4-y ^ 2)] dy + ∫ [(9 / 4) / (√ 9 / 4-y ^ 2)] dy this step is more critical, but not difficult
Then you get:
2*∫(√9/4-y^2)]dy=y(√9/4-y^2)+∫[(9/4)/(√9/4-y^2)]dy
How to integrate ∫ [(9 / 4) / (√ 9 / 4-y ^ 2)] dy?
In this way: ∫ [(9 / 4) / (√ 9 / 4-y ^ 2)] dy
=∫[(3/2)/(√1-(4/9)y^2)]dy
=(9/4)∫[1/(√1-(2y/3)^2)]d(2y/3)
=(9/4)arcsin(2y/3)+c
SO 2 * ∫ (√ 9 / 4-y ^ 2)] dy
=y(√9/4-y^2)+(9/4)arcsin(2y/3)+c
So ∫ (√ 2 + x-x ^ 2) DX
=(9/8)arcsin(2y/3)+(y/2)(√9/4-y^2)+c
=(9/8)arcsin(2x-1/3)+(x/2-1/4)(√2+x-x^2)+c
It's a little different from your answer, but I think there's something wrong with the answer,
Maybe I was wrong, but I think the process has been very detailed