Answer: [tex]8 \sin x +7(\dfrac{\sin^3x}{3})+C[/tex]
Step-by-step explanation:
Consider [tex]\int \cos(x) (8+7 \sin^2(x)) \, dx[/tex]
Substitute t= sinx
then dt = cos x dx
[tex]\int \cos(x) (8+7 \sin^2(x)) \, dx = \int (8+7t^2)dt\\\\ =8t+7(\dfrac{t^3}{3})+C[/tex]
[tex][\int x^ndx=\dfrac{x^{n+1}}{n+1}+C][/tex]
[tex]=8 \sin x +7(\dfrac{\sin^3x}{3})+C[/tex]
Hence, [tex]\int \cos(x) (8+7 \sin^2(x)) \, dx=8 \sin x +7(\dfrac{\sin^3x}{3})+C[/tex]
