<s>[INST] Integral by change of variable $\int_0^{\pi} \sin^4(x)\cos^6(x)\,dx $ 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....
 [/INST] $$\int_{0}^{\pi} \sin^4 x\cos^6 x dx =\int_{0}^{\pi} (\frac{1-\cos 2x}{2})^2 (\frac{1+\cos 2x}{2})^3 dx =$$
$$=\frac{1}{32}\int_{0}^{\pi} (1-\cos 2x)^2 (1+\cos 2x)^3 dx =$$
$$=\frac{1}{32}\int_{0}^{\pi} (1-2\cos 2x+\cos^2 2x) (1+3\cos 2x+3\cos^2 2x+\cos^3 2x) dx =$$
$$=\frac{1}{32}\int_{0}^{\pi} (1+\cos 2x-2\cos^2 2x-2\cos^3 2x+\cos^4 2x+\cos^5 2x) dx $$
$$\int_{0}^{\pi} (1) dx =\pi$$
$$\int_{0}^{\pi} (\cos 2x) dx =0$$
$$\int_{0}^{\pi} (\cos^2 2x) dx =\frac{\pi}{2}$$
$$\int_{0}^{\pi} (\cos^3 2x) dx =0$$
$$\int_{0}^{\pi} (\cos^4 2x) dx =\frac{3\pi}{8}$$
$$\int_{0}^{\pi} (\cos^5 2x) dx =0$$
$$=\frac{1}{32}(\pi -\pi+\frac{3\pi}{8})=\frac{3\pi}{256}$$
</s>