Fallacy - where is the mistake?

Question

Could anyone help me to find the mistake in this fallacy? Because the actual result for $I$ is $\pi/2$

\begin{equation} I = \int_{0}^{\pi} \cos^{2} x \; \textrm{d}x \end{equation}

\begin{equation} I = \int_{0}^{\pi} \cos x \cos x \; \textrm{d}x \end{equation} substitution: \begin{equation} \sin x = u \end{equation}

\begin{equation} \cos x \; \textrm{d}x = u \end{equation}

\begin{equation} \cos x = \cos (\arcsin x) \end{equation} and the limits: \begin{align} \begin{split} x &= 0 \quad \Rightarrow \quad u=0 \\ x &= \pi \quad \Rightarrow \quad u=0 \end{split} \end{align} so \begin{equation} I = \int_{0}^{0} \cos (\arcsin u) \textrm{d}u \end{equation}

\begin{equation} I=0 \end{equation}

Answer 1

The problem is in using $\cos x = \cos \sin^{-1} x$. Arcsine is not a single valued function on $x=[0,\pi]$. In the conventional sense, it's not defined outside of $[-\pi/2,\pi/2]$. You can only do a $u$ substitution using a one-to-one function.

Answer 2

if $\sin(x)=u$ we get $x=\arcsin(u)$ and $dx=\frac{1}{\sqrt{1-u^2}}du$ thus we get $\int \cos(x)^2dx=\int 1-\sin(x)^2dx=\int \sqrt{1-u^2}du$

Answer 3

You've messed something up...I can't quite see it, but this is how I would solve the problem:

\begin{equation} I=\int_0^\frac{\pi}{2}\cos^2\left(x\right)dx=\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{2}+\frac{\cos\left(2x\right)}{2}\right)dx \end{equation} via using the trigonometric identity. Your u-substitution is wrong btw...you cannot substitute $u=\sin\left(x\right)$ since there is no $\sin\left(x\right)$ in the integrand, and it isn't equivalent to anything either unless you change your bounds of integration which is pointless.