Answer
$-\dfrac{\cos^7 x \sin x}{8}+\dfrac{\cos^5 x \sin x}{48}-\dfrac{5\cos^3 x \sin x}{192}+\dfrac{ 5 \cos x \sin x}{128}+\dfrac{ 5 x}{128}+C $
Work Step by Step
In order to solve the integral, we will use the reduction formula:
$\int \cos^n x \ dx =\dfrac{1}{n} \cos^{n-1} x \ \sin x +\dfrac{(n-1) }{n}\int \cos^{n-2} x \ dx $
Now, $\int \sin^2 x \cos^6 x \ dx= \int (1-\cos^2 x) \cos^6 x \ dx \\=\int \cos^6 x dx -\int \cos^8 x \ dx$
Since, $\int \cos^8 x \ dx =\dfrac{1}{8} \cos^{7} x \ \sin x +\dfrac{7}{8}\int \cos^{6} x \ dx= \dfrac{\cos^7 x \sin x }{8}+\dfrac{7 \cos^5 x \sin x}{48}+\dfrac{35 \cos^3x \sin x}{192}+\dfrac{105 \cos x \sin x}{384}+\dfrac{105}{384} x+C$
Now, $I =\int \cos^6 x dx -\int \cos^8 x \ dx= \dfrac{\cos^5 x \sin x}{6}+\dfrac{ 5 \cos^3x \sin x}{24}+\dfrac{15 \cos x \sin x}{48}+\dfrac{15x}{48} -[\dfrac{\cos^7 x \sin x }{8}+\dfrac{7 \cos^5 x \sin x}{48}+\dfrac{35 \cos^3x \sin x}{192}+\dfrac{105 \cos x \sin x}{384}+\dfrac{105}{384} x]+C \\=\dfrac{\cos^5 x \sin x}{6}+\dfrac{ 5 \cos^3x \sin x}{24}+\dfrac{15 \cos x \sin x}{48}+\dfrac{15x}{48} -\dfrac{\cos^7 x \sin x }{8} -\dfrac{7 \cos^5 x \sin x}{48}-\dfrac{35 \cos^3x \sin x}{192}-\dfrac{105 \cos x \sin x}{384}-\dfrac{105}{384} x+C \\=-\dfrac{\cos^7 x \sin x}{8}+\dfrac{\cos^5 x \sin x}{48}-\dfrac{5\cos^3 x \sin x}{192}+\dfrac{ 5 \cos x \sin x}{128}+\dfrac{ 5 x}{128}+C $
