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By integrating by parts twice, show that $I_n$, as defined below for integers $n > 1$, has the value shown.

$$I_n = int_0^{pi / 2} sin n theta cos theta ,dtheta = frac{n-sin(frac{pi n}{2})}{n^2 -1}$$

I can do this using the formula $$sin A cos B = frac{1}{2}[sin(A-B)+sin(A+B)] ,$$ but when I try using integration by parts I get stuck in a loop of integrating the same thing over and over.

Hint Following what you've done already, integrating by parts with $u = sin n theta$, $dv = cos theta ,dtheta$ gives
begin{multline}color{#df0000}{I_n} = underbrace{sin n theta}_u , underbrace{sin theta}_v vert_0^{pi / 2} - int_0^{pi / 2} underbrace{sin theta}_v , underbrace{cos ntheta , dtheta}_{du} = sin frac{pi n}{2} - n color{#1f1fff}{J_n}, \ color{#1f1fff}{J_n := int_0^{pi / 2} cos n theta sin theta , dtheta} .end{multline}

We now apply integration by parts to the integral $color{#3f3fff}{J_n}$ with $p = cos n theta$, $dq = sin theta ,dtheta$:
$$color{#3f3fff}{J_n} = cos n theta (-cos theta)vert_0^{pi / 2} - int_0^{pi / 2} underbrace{-cos theta}_q cdot underbrace{- n sin n theta ,dtheta}_{dp} = 1 - n color{#df0000}{I_n} .$$

Substituting to eliminate $color{#3f3fff}{J_n}$ gives $color{#df0000}{I_n} = sin frac{pi n}{2} - n (1 - n color{#df0000}{I_n})$, and rearranging to solve for $color{#df0000}{I_n}$ gives the claimed identity: $$color{#df0000}{boxed{I_n = frac{n - sin frac{pi n}{2}}{n^2 - 1}}} .$$ Notice that for $n equiv 0, 2 pmod 4$ this simplifies to $frac{n}{n^2 - 1}$, for $n equiv 1 pmod 4$ to $frac{1}{n + 1}$, and for $n equiv 3 pmod 4$ to $frac{1}{n - 1}$.

$$cos{(x)} sin{left( n xright) }=frac{sin{left( left( n+1right) xright) }+sin{left( left( n-1right) xright) }}{2}$$
$$int{left. cos{(x)} sin{left( n xright) }dxright.}=-frac{cos{left( left( n+1right) xright) }}{2 left( n+1right) }-frac{cos{left( left( n-1right) xright) }}{2 left( n-1right) }$$
Then
$$int_{0}^{frac{pi}{2}}{left. cos{(x)} sin{left( n xright) }dxright.}\=
frac{n}{n^2-1}-frac{left( n-1right) cos{left( frac{{pi} n+{pi} }{2}right) }+left( n+1right) cos{left( frac{{pi} n-{pi} }{2}right) }}{2n^2-2}\=
frac{n}{{{n}^{2}}-1}-frac{sin{left( frac{{pi} n}{2}right) }}{{{n}^{2}}-1}$$

Hint:

There's an escape from the infinite loop. If you observe closely, you will see that after two iterations you get to something like

$$I=b+aI$$

where the constants $a,b$ are computable.

You naturally conclude

$$I=frac{b}{1-a}.$$

$$intsin nthetacostheta,dtheta=sin nthetasintheta-nintcos nthetasintheta,dtheta\=sin nthetasintheta+ncos nthetacostheta+n^2intsin nthetacostheta,dtheta.$$