Trig factoring

**Greymalkin** · November 19th 2012, 05:25 PM

Through what operation(s) does

$\frac{1}{5}[sin(4x)+1]^\frac{5}{2}-\frac{1}{14}[sin(4x)+1]^\frac{7}{2}+C$

become

$\frac{1}{70}[9-5sin(4x))(sin(4x)+1]^\frac{5}{2}+C$

**MarkFL** · November 19th 2012, 05:50 PM

Observe that the initial expression may be written:

$\frac{14}{70}\left(\sin(4x)+1 \right)^{\frac{5}{2}}-\frac{5}{70}\left(\sin(4x)+1 \right)\left(\sin(4x)+1 \right)^{\frac{5}{2}}+C$

Now factor:

$\frac{1}{70}\left(\sin(4x)+1 \right)^{\frac{5}{2}}\left(14-5(\sin(4x)+1) \right)+C=$

$\frac{1}{70}\left(9-5\sin(4x) \right)\left(\sin(4x)+1 \right)^{\frac{5}{2}}+C$

**Greymalkin** · November 19th 2012, 06:48 PM

Originally Posted by MarkFL2

Observe that the initial expression may be written:

$\frac{14}{70}\left(\sin(4x)+1 \right)^{\frac{5}{2}}-\frac{5}{70}\left(\sin(4x)+1 \right)\left(\sin(4x)+1 \right)^{\frac{5}{2}}+C$

I understand that you remove a factor of sin4x+1 from the exponent, but where do the $\frac{14}{70}$ and the $\frac{5}{70}$ come from?

Now factor:

Originally Posted by MarkFL2

$\frac{1}{70}\left(\sin(4x)+1 \right)^{\frac{5}{2}}\left(14-5(\sin(4x)+1) \right)+C=$

$\frac{1}{70}\left(9-5\sin(4x) \right)\left(\sin(4x)+1 \right)^{\frac{5}{2}}+C$

What are you doing with the 14 and the 9?

**MarkFL** · November 19th 2012, 06:56 PM

I am rewriting the given fractions using a common denominator of 70.

Once you factor the 1/70 out, then you are left with the numerators of 14 and 5. From there it is a matter of distributing and collecting like terms.

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