Question

Determine integration
1) (cos 2x + sin4x) dx
2) (sin 4xcos2x-cos 4xsin2x)
3) (cos⁴x-sin⁴x) dx

Answer 1

$\displaystyle \int\cos^4x-\sin^4x\,dx=\int(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)\,dx\\\int\cos^4x-\sin^4x\,dx=\int(\cos2x)(1)\,dx\\\int\cos^4x-\sin^4x\,dx=\int\cos2x\,dx\\\boxed{\boxed{\int\cos^4x-\sin^4x\,dx=\frac12\sin2x+C}}$
$\displaystyle \int\sin 4x\cos2x-\cos 4x\sin2x\,dx=\int\sin(4x-2x)\,dx\\\int\sin 4x\cos2x-\cos 4x\sin2x\,dx=\int\sin2x\,dx\\\boxed{\boxed{\int\sin 4x\cos2x-\cos 4x\sin2x\,dx=-\frac12\cos2x+C}}$
$\displaystyle \int\cos2x+\sin^4x\,dx=\int\cos2x\,dx+\int\sin^4x\,dx\\\int\cos2x+\sin^4x\,dx=\frac12\sin2x+\frac{4-1}{4}\int\sin^{4-2}x\,dx+\frac{\cos x\sin^{4-1}x}{4}\\\int\cos2x+\sin^4x\,dx=\frac12\sin2x+\frac{3}{4}\int\sin^{2}x\,dx+\frac{\cos x\sin^{3}x}{4}\\\int\cos2x+\sin^4x\,dx=\frac12\sin2x+\frac{3}{4}\int\frac{1-\cos2x}{2}\,dx+\frac{\cos x\sin^{3}x}{4}\\\int\cos2x+\sin^4x\,dx=\frac12\sin2x+\frac{3}{4}\cdot\frac{x-\frac12\sin2x}{2}+\frac{\cos x\sin^{3}x}{4}+C\displaystyle \int\cos2x+\sin^4x\,dx=\frac12\sin2x+\frac{6x-3\sin2x}{16}+\frac{\cos x\sin^{3}x}{2}+C\\\int\cos2x+\sin^4x\,dx=\frac12\sin2x+\frac{3x}{8}+\frac{8\cos x\sin^{3}x-2\sin2x-\sin2x}{16}+C$$\displaystyle \int\cos2x+\sin^4x\,dx=\frac12\sin2x+\frac{3x}{8}+\frac{\sin4x-\sin2x}{16}+C\\\int\cos2x+\sin^4x\,dx=\frac{3x}{8}+\frac{8\sin2x+\sin4x-\sin2x}{16}+C\\\boxed{\boxed{\int\cos2x+\sin^4x\,dx=\frac{3x}{8}+\frac{7\sin2x+\sin4x}{16}+C}}$