Integration of inverse trigonometric functions

[delapcsoncruz]

Homework Statement

∫ (x+2)dx/√(4x-x²)

Homework Equations

why was the -2 in -2(x-2) was ignored?

The Attempt at a Solution

so first i let u= 4x-x²
then, du=4-2x
= -2(x-2)

so to get (x+2) i equate it to (x-2)+4

so ...
∫ (x+2)dx/√(4x-x²) = ∫(x-2)+4dx/√(4x-x²)

= ∫ (x-2)dx/√(4x-x²) + ∫4dx/√(4x-x²)is the solution right?
and why was the -2 in -2(x-2) was ignored?

 

[SammyS]

Homework Statement

∫ (x+2)dx/√(4x-x²)

Homework Equations

why was the -2 in -2(x-2) was ignored?

The Attempt at a Solution

so first i let u= 4x-4x²
then, du=4-2x
= -2(x-2)

so to get (x+2) i equate it to (x-2)+4

so ...
∫ (x+2)dx/√(4x-x²) = ∫(x-2)+4dx/√(4x-x²)

= ∫ (x-2)dx/√(4x-x²) + ∫4dx/√(4x-x²)

is the solution right?
and why was the -2 in -2(x-2) was ignored?

Hello delapcsoncruz. Welcome to PF !

Check your differentiation.

du = (4-8x)dx​

As to your question: "why was the -2 in -2(x-2) was ignored?" : What are you referring to ?

 

[delapcsoncruz]

Hello delapcsoncruz. Welcome to PF !

Check your differentiation.

du = (4-8x)dx​

As to your question: "why was the -2 in -2(x-2) was ignored?" : What are you referring to ?

sorry typographical error.. the u= 4x-x²
so du = 4-2x = -2(x-2)

 

[SammyS]

sorry typographical error.. the u= 4x-x²
so du = 4-2x = -2(x-2)

Not quite right.

du = -2(x-2)dx .

You're doing the method of substitution incorrectly. At any rate, the substitution you're trying won't help much.

Complete the square for -x² + 4x .

You should then be able to let u = x-2 .

 

[delapcsoncruz]

Not quite right.

du = -2(x-2)dx .

You're doing the method of substitution incorrectly. At any rate, the substitution you're trying won't help much.

Complete the square for -x² + 4x .

You should then be able to let u = x-2 .

thank you...
so ∫ (x+2)dx / √ (4x-x²) = ∫ (x+2)dx / √ (4-(x+2)²)

i don't know what is the next thing to do.. please help..

 

[Curious3141]

thank you...
so ∫ (x+2)dx / √ (4x-x²) = ∫ (x+2)dx / √ (4-(x+2)²)

i don't know what is the next thing to do.. please help..

The completion of the square has not been done properly. $\sqrt{4x-x^2} \neq \sqrt{4 - {(x+2)}^2}$. Remember that you have to get a $-4x$ term from the square expression to end up with a (positive) 4x at the end.

 

[delapcsoncruz]

The completion of the square has not been done properly. $\sqrt{4x-x^2} \neq \sqrt{4 - {(x+2)}^2}$.

thank you for the correction...
sorry its ∫ (x+2)dx / √ 4 - (x-2)²

so what i am going to do next?

 

[Curious3141]

thank you for the correction...
sorry its ∫ (x+2)dx / √ 4 - (x-2)²

so what i am going to do next?

Ah good, you got it. You didn't need the hint I added in my edit.

OK, now you can do the sub. that Sammy was suggesting, then separate the integrand into two simpler quotients added together (separate the sum in the numerator). After that, it should become quite easy to see.

 

[delapcsoncruz]

Ah good, you got it. You didn't need the hint I added in my edit.

OK, now you can do the sub. that Sammy was suggesting, then separate the integrand into two simpler quotients added together (separate the sum in the numerator). After that, it should become quite easy to see.

how can i do the substitution? can you give me a hint.

 

[Curious3141]

how can i do the substitution? can you give me a hint.

Start by subbing u = (x-2).

What does the numerator become in terms of u? Denominator? What about dx in terms of du?

 

[delapcsoncruz]

Start by subbing u = (x-2).

What does the numerator become in terms of u? Denominator? What about dx in terms of du?

i solve it and this is my solution..

∫ (x+2)dx / sqrt (4x-x²)
let u= 4x -x²
du=-2(x-2)dx

so now the numerator will become [(x-2)+4]dx

= ∫ [(x-2)+4]dx / sqrt (4x-x²)

= ∫ (x-2)dx / sqrt (4x-x²) + ∫ 4dx / sqrt (4x-x²)

let u = 4x -x²
du = -2(x-2)dx

by completing the square : 4x - x² = 4-(x-2)²

= -1/2 ∫ du / u^1/2 + 4∫ dx / sqrt 4 -(x-2)²
= -u^1/2 + 4∫ du / sqrt (a² -u²)

= -√(4x -x²) + 4 Arcsin (x-2)/2 +c


is the answer right?

 

[Curious3141]

i solve it and this is my solution..

∫ (x+2)dx / sqrt (4x-x²)
let u= 4x -x²
du=-2(x-2)dx

so now the numerator will become [(x-2)+4]dx

= ∫ [(x-2)+4]dx / sqrt (4x-x²)

= ∫ (x-2)dx / sqrt (4x-x²) + ∫ 4dx / sqrt (4x-x²)

let u = 4x -x²
du = -2(x-2)dx

by completing the square : 4x - x² = 4-(x-2)²

= -1/2 ∫ du / u^1/2 + 4∫ dx / sqrt 4 -(x-2)²
= -u^1/2 + 4∫ du / sqrt (a² -u²)

= -√(4x -x²) + 4 Arcsin (x-2)/2 +c


is the answer right?

Your final answer is right (and you decided to take a slightly more complicated route, but that's OK).

However, some of your notation needs cleaning up (it would be grounds for marking you wrong or subtracting marks). For example, in one line, "u" has two different definitions in the two integrals. And then "a" is brought in out of nowhere (you were probably referencing a standard form, but still, the "a" doesn't belong in the middle of your working).

 

[delapcsoncruz]

Your final answer is right (and you decided to take a slightly more complicated route, but that's OK).

However, some of your notation needs cleaning up (it would be grounds for marking you wrong or subtracting marks). For example, in one line, "u" has two different definitions in the two integrals. And then "a" is brought in out of nowhere (you were probably referencing a standard form, but still, the "a" doesn't belong in the middle of your working).

thank you very much! ok i will revise my solution to make it more understandable... thank you again..