# Solve integral $\int_0^\pi \sin(t-2nt)dt$

I have the integral $$\int_0^\pi \sin(t-2nt)dt$$

Wolfram states that the answer is: $$\frac{2\cos^2(n\pi)}{1-2n}$$ But I can't get the same... I am close to the answer with the calculation below, but something is not right:

$$\int_0^\pi \sin(t-2nt)dt = \begin{Bmatrix} u=t(1-2n) \\ du = 1-2n dt\end{Bmatrix} = \int_0^\pi \sin u\frac{1}{1-2n}du=\begin{bmatrix}\frac{-1}{1-2n}\cos(t(1-2n))\end{bmatrix}_0^\pi = \frac{-1}{1-2n}\cos(\pi-2n\pi)=\\=\frac{-1}{1-2n}\cos\pi\cos(2n\pi)+\sin\pi\sin(2n\pi) = \frac{\cos(2n\pi)}{1-2n}$$

Where does it fail?

• When you computed the value at the bounds, you forgot to substract the value for $t=0$ (which is not $0$!).Fix it and remember that $\cos(2x)=2\cos^2(x)-1$. You are almost done. Commented Sep 15, 2014 at 8:02
• Also, watch out for the case $n=\frac12$. (Though your $n$ may be intended to be integer.) Commented Sep 15, 2014 at 8:28

You haven't substituted the lower limit of integration which is $0$. The value of the expression after you substitue lower limit is not zero as you will see when you simplify it carefully.
$$\int_0^{\pi}\sin(t-2nt)dt\\=\int_0^{\pi}\sin ((1-2n)t) dt\\=-\left[\dfrac{\cos(1-2n)t}{(1-2n)}\right]^\pi_0\\=\frac{1}{1-2n}-\frac{1}{1-2n}\cos (1-2n)\pi \\ =\frac{1-\cos 2n\pi}{1-2n}\\=\frac{2\cos^2n\pi}{1-2n} \blacksquare$$
Let $(1-2n)t=u$ so we have $dt=\frac{1}{1-2n}du$ and so $$\int_{0}^{\pi}\sin(t-2nt)\, dt=\int_{0}^{\pi} \, \frac{\sin u}{1-2n}du=\frac{1}{1-2n}\int_{0}^{\pi} \sin u\,du$$ therefore we see that
\begin{align} \int_{0}^{\pi}\sin(t-2nt)\, dt& =\left[\frac{-1}{1-2n}\cos u \right]_{0}^{\pi}\\ & =\left[\frac{-1}{1-2n}\cos (1-2n)t \right]_{0}^{\pi}\\ &\stackrel{\color {\red}1}=\frac{1}{1-2n}-\frac{1}{1-2n}\cos (1-2n)\pi\\&=\frac{1-\cos 2n\pi}{1-2n}\\& =\frac{2\cos^2 n\pi}{1-2n} \end{align} $\cos^2 x=\frac{1-\cos 2x}{2}$ and you do not substitute $0$ in $(1)$