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.
 A: Let $$\displaystyle I = \int (5-5\sin x)\cdot (5+5\sin x)dx = 25\int (1-\sin^2 x)dx = 25\int \cos^2 xdx$$
Now Using Integration by parts,
So $$\displaystyle I = 25\int \cos x\cdot \cos xdx = 25\cos x\cdot \sin x+25\int \sin x\cdot \sin xdx$$
Using $$\bullet\; \sin 2x = 2\sin x\cdot \cos x$$
So $$\displaystyle I = \frac{25}{2}\sin 2x+25\int (1-\cos^2 x)dx = \frac{25}{2}\sin 2x+25x-I$$
so we get $$\displaystyle I = \frac{25}{4}\sin 2x+\frac{25}{2}+\mathcal{C}$$
A: 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,
$$I = \frac{25}{4} \, \left[ 2 \, x + \sin(2 \, x) \right]$$
A: However if you really want integration by parts: 
\begin{align}
\int \sin(x) \cdot \sin(x) dx &= \sin(x) \cdot (-\cos(x))-\int \cos(x) \cdot (-\cos(x) )dx+C \\
&= -\sin(x) \cdot \cos(x)+\int \cos^2(x) dx +C \\
&= -\sin(x) \cdot \cos(x)+\int(1-\sin^2(x) )dx +C \\
&= -\sin(x) \cos(x)+x -\int \sin^2(x) dx +C.
\end{align} 
And so add 
$$\int \sin^2(x) dx $$ 
on both sides.. 
$$2 \int \sin^2(x) dx=-\sin(x) \cos(x)+x+C$$
