integral of sec^3x 0.5(sec2x)(sinx) - 0.5∫secx dx: how?

[hndalama]

integral of sec^3x 0.5(sec2x)(sinx) - 0.5∫secx dx: how?

in solving this integral my book says that ∫sec³x dx can be written as = 0.5(sec²x)(sinx) - 0.5∫secx dx
Can anyone explain how they derive that?

 

[MarkFL]

Let's look at what is said by your book to be an identity:

\(\displaystyle \displaystyle \int \sec^3(x)\,dx=\frac{1}{2}\sec^2(x)\sin(x)-\frac{1}{2}\int \sec(x)\,dx\)

If we differentiate with respect to \(\displaystyle x\), we obtain:

\(\displaystyle \displaystyle \sec^3(x)=\frac{1}{2}\left(\sec^2(x)\cos(x)+ 2\tan(x)\sec^2(x)\sin(x)\right)-\frac{1}{2}\sec(x)\)

\(\displaystyle \displaystyle \sec^3(x)=\sin^2(x)\sec(x)\)

This isn't an identity...but let's try:

\(\displaystyle \displaystyle \int \sec^3(x)\,dx=\frac{1}{2}\sec^2(x)\sin(x)+\frac{1}{2}\int \sec(x)\,dx\)

\(\displaystyle \displaystyle \sec^3(x)=\frac{\sec(x)}{2}\left(\tan^2(x)+\sec^2(x)+1\right)\)

\(\displaystyle \displaystyle 2\sec^2(x)=\tan^2(x)+\sec^2(x)+1\)

\(\displaystyle \displaystyle \sec^2(x)=\tan^2(x)+1\quad\checkmark\)

This is a Pythagorean identity.

 

[hndalama]

How do you get from:
\(\displaystyle \displaystyle \sec^3(x)=\frac{1}{2}\left(\sec^2(x)\cos(x)+ 2\tan(x)\sec^2(x)\sin(x)\right)-\frac{1}{2}\sec(x)\)

to \(\displaystyle \displaystyle \sec^3(x)=\sin^2(x)\sec(x)\)

I end up with
\(\displaystyle \displaystyle \sec^3(x)=\tan^2(x)\sec(x)\)

 

[MarkFL]

How do you get from:
\(\displaystyle \displaystyle \sec^3(x)=\frac{1}{2}\left(\sec^2(x)\cos(x)+ 2\tan(x)\sec^2(x)\sin(x)\right)-\frac{1}{2}\sec(x)\)

to \(\displaystyle \displaystyle \sec^3(x)=\sin^2(x)\sec(x)\)

I end up with
\(\displaystyle \displaystyle \sec^3(x)=\tan^2(x)\sec(x)\)

Yes, I made an error there...I should have written:

\(\displaystyle \displaystyle \sec^3(x)=\sin^2(x)\sec^3(x)\)

Which is equivalent to what you correctly obtained.