<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] $$\sin^4x\cos^6x=\frac{1}{16}\cos^2x\sin^42x=\frac{1}{128}(1+\cos2x)(1-\cos4x)^2=$$
$$=\frac{1}{128}(1+\cos2x)\left(1-2\cos4x+\frac{1+\cos8x}{2}\right)=$$
$$=\frac{1}{256}(1+\cos2x)(3-4\cos4x+\cos8x),$$
which says that our integral it's
$$\int\limits_0^{\pi}\frac{3}{256}dx=\frac{3\pi}{256}.$$
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