mathematics help on how i can get these answers and what i did wrong?

Question

any tips would be great.


1. antideriv of x(4thrt(1-x^2))dx
i got -2/5(1-x^2)^((5/4))+c but the book says it should be ^(5/2)

2. antideriv of (sin(x/3))^5(cos(x/3))dx
i got -1/6(sinx/3)^6+c but that book says its 1/2 in the front

3. antideriv of sec(x+pi/2)tan(x+pi/2)dx
i got cos(x+pi/2)+c but the book says its secant

4. antideriv of (sin(2x+1))/(cos(2x+1))^2dx
i got cos(2x+1)^-1 but the book says there is a 1/2 in front

5. antideriv of (sqrt(cotx))(cscx)^2dx
couldnt get much anywhere in this one

6. antideriv of (1/x^2)sin(1/x)cos(1/x)dx
i think this one is all wrong cuz i got -sin(1/x)+c and the book has it sqrd and times 1/2

7. antideriv of (x^3)(sqrt(x^2+1))
i dont know where to start because what i thought would be u is not able to be because of the x^3... its one x too big!!

Answer 1

I see!
You do not need to be panic. However, be VERY careful.
Let me use [int] to represent the integral sign. Also, I would like to show you a way NOT to use u, v, etc. to make you confuse. But, please pay close attention to the item after d, such as x in dx, or x^2 in d(x^2)

1) [int] x (1-x^2)^(1/4)dx
= (1/2) [int] (1-x^2)^(1/4)d(x^2)
= -(1/2) [int] (1-x^2)^(1/4)d(1 - x^2)
= -(1/2)(4/5) (1-x^2)^(5/4) + C
Your answer is correct.

2) [int] (sin(x/3))^5(cos(x/3))dx
= 3 [int] (sin(x/3))^5(cos(x/3))d(x/3)
= 3 [int] (sin(x/3))^5d[sin(x/3)]
= (1/6)3(sin(x/3))^6 +C

3) [int] sec(x+pi/2)tan(x+pi/2)dx
= [int] sin(x+pi/2) d(x+pi/2) / cos^2 (x+pi/2)
= -[int] d[cos(x+pi/2)] / cos^2 (x+pi/2)
= 1/cos(x+pi/2) + C

4) [int] (sin(2x+1))/(cos(2x+1))^2dx
= (1/2) [int] (sin(2x+1))/(cos(2x+1))^2d(2x+1)
Now you see where (1/2) comes from.

5) [int] (sqrt(cot x))(csc x)^2dx
= -[int] (sqrt(cot x))d(cot x)
-- you should know d(cot x)/dx = - (csc x)^2

6) [int] (1/x^2)sin(1/x)cos(1/x)dx
= -[int] sin(1/x)cos(1/x)d(1/x)
= -[int] sin(1/x)d[sin(1/x)]
Now do you know where you made mistakes?

7) [int] (x^3)(sqrt(x^2+1))dx
= (1/2) [int] (x^2)(sqrt(x^2+1))(2x)dx
= (1/2) [int] (x^2)(sqrt(x^2+1))d(x^2)
= (1/2) [int] (x^2)(sqrt(x^2+1))d(x^2+1)
= (1/2)(2/3) [int] (x^2)d[(x^2+1)^(3/2)]

= (1/3)(x^2)*(x^2+1)^(3/2) - (1/3) [int] (x^2+1)^(3/2)d(x^2)
= (1/3)(x^2)*(x^2+1)^(3/2) - (1/3) [int] (x^2+1)^(3/2)d(x^2+1)
you can do it now!