Evaluate the antiderivatives \[ \int\sin^2x\,dx,\quad \int\cos^2x\,dx,\quad \int\sin^3x\,dx,\quad \int\cos^3x\,dx, \] \[ \int(1+\sin x)^3\,dx,\quad \int(1+\cos x)^3\,dx, \] \[ \int\sin^4x\,dx,\quad \int\cos^4x\,dx, \quad \int(\cos^4x-\sin^4x)\,dx, \] and \[ \int \frac{\sin^3\sqrt x}{2\sqrt x}\,dx. \]
Solution
Using the power-reduction identities and elementary substitutions, \[ \boxed{\int\sin^2x\,dx=\frac12(x-\sin x\cos x)+C,} \] \[ \boxed{\int\cos^2x\,dx=\frac12(x+\sin x\cos x)+C,} \] \[ \boxed{\int\sin^3x\,dx=\frac13\cos^3x-\cos x+C,} \] \[ \boxed{\int\cos^3x\,dx=\sin x-\frac13\sin^3x+C.} \] With \(u=\sqrt x\), the nested-radical variant is \[ \boxed{\int \frac{\sin^3\sqrt x}{2\sqrt x}\,dx =\frac13\cos^3\sqrt x-\cos\sqrt x+C.} \] Expanding the cubes gives \[ \boxed{\int(1+\sin x)^3dx =\frac52x+\frac13\cos^3x-4\cos x-\frac34\sin2x+C,} \] \[ \boxed{\int(1+\cos x)^3dx =\frac52x-\frac13\sin^3x+4\sin x+\frac34\sin2x+C.} \] Reduction formulas yield \[ \boxed{\int\sin^4x\,dx =\frac38x-\frac38\sin x\cos x-\frac14\sin^3x\cos x+C,} \] \[ \boxed{\int\cos^4x\,dx =\frac38x+\frac38\sin x\cos x+\frac14\sin x\cos^3x+C.} \] Finally \(\cos^4x-\sin^4x=\cos2x\), so \[ \boxed{\int(\cos^4x-\sin^4x)dx=\frac12\sin2x+C.} \]