91Ó°ÊÓ

Rewrite the integral as \(\int \sin x\cdot\sin^2 x\, dx\).

Use the identity \(\sin^2 x = 1 - \cos^2 x\) to rewrite the integral as \(\int \sin x (1 - \cos^2 x)\, dx\).

Integration by parts states that \(\int u dv = uv - \int v du\). Let's choose \(u = \sin x\) and \(dv = (1 - \cos^2 x) dx\). Then, we have: $$ du = \cos x \, dx \\ v = \int (1 - \cos^2 x) dx = x - \int \cos^2 x\, dx $$ Now, the integral becomes: $$ \int \sin x (1 - \cos^2 x)\, dx = \sin x (x - \int \cos^2 x\, dx) - \int \cos x (x - \int \cos^2 x\, dx) dx $$

The integral of \(\cos^2 x \) can be found using another trigonometric identity and half-angle formula to make the integral: $$ \int \cos^2 x\, dx = \frac{1}{2}\int (1 + \cos 2x) dx = \frac{1}{2}(x+\frac{\sin 2x}{4}) + C_1 $$ And the remaining integral is: $$ \int \cos x (x - \int \cos^2 x\, dx) dx = \int x\cos x\, dx - \frac{1}{2}\int \cos x(\cos 2x) dx $$ Now, applying integration by parts to \(\int x\cos x\, dx\), and using substitution for the second integral, we get: $$ \int x\cos x\, dx = x\int \cos x\, dx - \int (\int \cos x\, dx) dx = x\sin x - \int \sin x\, dx = x\sin x + \cos x + C_2 $$ And for the second integral, we have: $$ \frac{1}{2}\int \cos x(\cos 2x) dx = \frac{1}{2}\int \cos x\frac{1+\cos 2x}{2} dx = \frac{1}{4}\int (1+\cos 2x) \cos x\, dx $$ Applying substitution \(u=2x\) and taking the integral, we end up with: $$ \frac{1}{4}\int (1+\cos 2x) \cos x\, dx = \frac{1}{4}(\frac{x}{2} + \frac{1}{4}\sin 2x) + C_3 $$ Combining all the results, we have: $$ \int \sin^3 x\, dx = \sin x (x - \frac{1}{2}(x+\frac{\sin 2x}{4})) - (x\sin x + \cos x + \frac{1}{8}x - \frac{1}{16}\sin 2x) + C $$ Simplify the expression to get the final result: $$ \int \sin^3 x\, dx = \frac{3}{8} x\sin x - \frac{1}{4}\sin x \cos x + \frac{1}{16}\cos x \sin 2x + C $$