Apostol's Theorem 2.5, sine cosine example

[hotwater]

Homework Statement

In Example 3 under Theorem 2.5 of Apostol's Calculus Volume 1, I don't understand how the final formula on the right is obtained.

Homework Equations

The identity cos 2x = 1 - 2 sin² x implies sin²x = 1/2(1 - cos 2x) so, from Example 2, we obtain:
\int^a_0\,sin^2\,x\,dx = \frac{1}{2}\,\int^a_0(1 - cos\,2x)\,dx = \frac{a}{2} - \frac{1}{4}\,sin\,2a

The Attempt at a Solution

I guess this uses (2.14), but I don't understand how:
(2.14): \int^{a}_{0} sin x dx = 1 - cos a

Scribd has a copy of the book online, at http://www.scribd.com/doc/36116063/eBook-Tom-Apostol-Calculus#. My question is in regard to page 101, Example 3, near the bottom.

 

[Apphysicist]

Homework Statement

In Example 3 under Theorem 2.5 of Apostol's Calculus Volume 1, I don't understand how the final formula on the right is obtained.

Homework Equations

The identity cos 2x = 1 - 2 sin² x implies sin²x = 1/2(1 - cos 2x) so, from Example 2, we obtain:
\int^a_0\,sin^2\,x\,dx = \frac{1}{2}\,\int^a_0(1 - cos\,2x)\,dx = \frac{a}{2} - \frac{1}{4}\,sin\,2a

The Attempt at a Solution

I guess this uses (2.14), but I don't understand how:
(2.14): \int^{a}_{0} sin x dx = 1 - cos a

Scribd has a copy of the book online, at http://www.scribd.com/doc/36116063/eBook-Tom-Apostol-Calculus#. My question is in regard to page 101, Example 3, near the bottom.

They used the trig identity to replace sin(x)^2 with (1-cos(2x))/2...then pulled the 1/2 out of the integral and integrated each term. Integral of 1 = x. They did a u-sub on cos(2x) with u=2x, du=2dx, thus it gets another 1/2 out front, and integral of sin = cos.