How Is the Book's Answer 2arcsec(√x) Derived in This Calc II Problem?

[Agent M27]

Homework Statement

$$\int\frac{dx}{\sqrt{x}\sqrt{1-x}}$$

Homework Equations

$$\int\frac{du}{\sqrt{a^{2}-u^{2}}}= arcsin\frac{u}{a} + C$$

The Attempt at a Solution

u$$^{2}$$=x

dx=2u du

$$\int\frac{2u}{u\sqrt{1-u^{2}}}du$$

2$$\int\frac{du}{\sqrt{1-u^{2}}} = 2arcsin\frac{u}{a} + C$$

=2arcsin($$\sqrt{x}$$) + C

But the book gives the answer to be 2arcsec($$\sqrt{x}$$) + C, which I do not understand how they achieved that answer. Any help would be appreciated. Thank you.

Joe

 

[NastyAccident]

Homework Statement

$$\int\frac{dx}{\sqrt{x}\sqrt{1-x}}$$

Homework Equations

$$\int\frac{du}{\sqrt{a^{2}-u^{2}}}= arcsin\frac{u}{a} + C$$

The Attempt at a Solution

u$$^{2}$$=x

dx=2u du

$$\int\frac{2u}{u\sqrt{1-u^{2}}}du$$

2$$\int\frac{du}{\sqrt{1-u^{2}}} = 2arcsin\frac{u}{a} + C$$

=2arcsin($$\sqrt{x}$$) + C

But the book gives the answer to be 2arcsec($$\sqrt{x}$$) + C, which I do not understand how they achieved that answer. Any help would be appreciated. Thank you.

Joe

Well, rest easy.

I just checked your work and it is definitely $$2arcsin(sqrt(x)) + C$$. The answer that was given by the book is an unfortunate typo.

Also, u = sqrt(x) is another way to use substitution.

 

[Agent M27]

That is good to hear, I felt like I was going crazy. BTW I originally set u=sqrt(x), but when finding du, it was simpler to square both sides. I don't like radicals... Thank you.

Joe