## Calculus (3rd Edition)

$\dfrac{x}{16} +\dfrac{\sin^5 x \cos x}{6}-\dfrac{\sin^3 x \cos x}{24}-\dfrac{\sin x \cos x }{16}+ C$
In order to solve the integral, we will use the reduction formula: $\int \sin^n x \ dx =\dfrac{-1}{n} \sin^{n-1} x \ \cos x +\dfrac{(n-1) }{n}\int \sin^{n-2} x \ dx$ Now, $\int \sin^4 x \cos^2 x \ dx= \int \sin^4 x (1-\sin^2 x) \ dx \\=\int \sin^4 x dx -\int \sin^6 x \ dx$ Since, $\int \sin^6 x \ dx =\dfrac{-1}{6} \sin^{5} x \ \cos x +\dfrac{5}{6}\int \sin^{4} x \ dx= \dfrac{15 x}{48}-\dfrac{\sin^5 x \cos x}{6}-\dfrac{5 \sin^3x \cos x}{24}-\dfrac{15 \sin x \cos x}{48} +C$ So, $I=\int \sin^4 x -\int \sin^6 x \ dx = \dfrac{3x}{8}-\dfrac{\sin^3x \cos x}{4}-\dfrac{3 \sin x \cos x}{8}-[\dfrac{15 x}{48}-\dfrac{\sin^5 x \cos x}{6}-\dfrac{5 \sin^3x \cos x}{24}-\dfrac{15 \sin x \cos x}{48}]+C \\=\dfrac{3x}{8}-\dfrac{\sin^3x \cos x}{4}-\dfrac{3 \sin x \cos x}{8}-\dfrac{15 x}{48}+ \dfrac{\sin^5 x \cos x}{6} +\dfrac{5 \sin^3x \cos x}{24} +\dfrac{15 \sin x \cos x}{48}+C\\=\dfrac{x}{16} +\dfrac{\sin^5 x \cos x}{6}-\dfrac{\sin^3 x \cos x}{24}-\dfrac{\sin x \cos x }{16}+ C$