integration by parts of $25\, (1-\sin^{2}x)$ I need help solving this integration of parts problem. I've tried a few different solutions and keep getting the wrong answer. This question is in regards to this problem take the integral by parts of: 
$$\displaystyle \int (5-5\sin x)(5+5\sin x)dx$$
So first I multiply, and get $25-25\sin^2 x.$ 
Then i tried to use the formula, integral 
$$\displaystyle (f(x)g'(x)) = f(x)g(x) - \int (f'(x)g(x)).$$ 
But it was to no avail. I know the answer, if you'd like it provided 
but obviously more important is the how! Please help me and thanks.

Consider 
\begin{align}
I &= \int (5 - 5 \sin x) \cdot (5 + 5 \sin x) \, dx \\
&= 25 \, \int (1-\sin x) \cdot (1 + \sin x) \, dx = 25 \, J
\end{align}
Let $dv = 1-\sin x$ then $v = x + \cos x$, $u=1+\sin x$, $du = \cos x$ and
\begin{align}
J &= (x+\cos x)(1 + \sin x) - \int (x + \cos x) \, \cos x \, dx \\
&= (x + \cos x)(1 + \sin x) - \int x \, \cos x \, dx - \int \cos^{2}x \, dx \\
&= (x + \cos x)(1 + \sin x) - x \, \sin x - \cos x - J \\
\end{align}
or 
\begin{align}
J &= \frac{1}{2} \left[ (x + \cos x) (1 + \sin x) - x \, \sin x - \cos x \right] \\
&= \frac{1}{2} \left( x + \sin x \, \cos x \right) = \frac{2x + \sin(2x)}{4}
\end{align}
Now,
$$I = \frac{25}{4} \, \left[ 2 \, x + \sin(2 \, x) \right]$$
