TITLE: Integral by change of variable $\int_0^{\pi} \sin^4(x)\cos^6(x)\,dx $
QUESTION [5 upvotes]: What is $$\int_0^{\pi} \sin^4(x)\cos^6(x)\,dx $$
Putting $\sin(x) = t$, then $\cos(x) = \sqrt{1-t^2}$ and $\cos(x)\,dx = dt$
At $x = 0$, $\sin(x) = 0$, $\therefore\ t = 0$
At $x = \pi$, $\sin(x) = 0$, $\therefore\ t = 0$
Integral becomes
$$\int_0^0 t^4\cdot\sqrt[5]{1-t^2}\,dt = 0$$
Integral should not evaluate to zero
What is going on here, i cannot find any mistake....

REPLY [3 votes]: Others have already presented several approaches to find this integral, but let me offer a very short answer to your actual question:

What is going on here, i cannot find any mistake....

Your main mistake is here:

Putting $\sin(x)=t$ then $\color{red}{\cos(x)=\sqrt{1−t^2}}$.

The statement in red is not true. Since the integral ranges from $x=0$ to $x=\pi$, you have values of $x$ in quadrants I and II. But in quadrant II, i.e. for $x\in(\frac{\pi}{2},\pi)$, the values of $\cos(x)$ are negative. In other words, for $x\in(\frac{\pi}{2},\pi)$, we have
$$\cos(x)=-\sqrt{1−t^2}\neq\sqrt{1−t^2},$$
invalidating your proposed substitution.