How to Integrate $ \int \sin^2x\cos\ x \ dx $ without substitution

Question

How can I calculate the following integral without using substitution?

$$ \int \sin^2x\,\cos\ x \, dx $$

I have been stuck on this problem for about a day and cannot seem to come to a conclusion.

Answer 1

Identify the integral as one of the form

$$\int f'f^n\,dx=\frac{f^{n+1}}{n+1}+C$$

This is an-almost immediate integral.

Answer 2

Using the equality $\cos x=D_x(\sin\, x)$ we can see that the integral $\int \sin^2x\cdot\cos x \, dx$ is equal to integral $\int \sin^2x \cdot D_x(\sin\, x) \, dx$. That is $$ \int \sin^2x\cdot\cos x \, dx = \int \sin^2x \cdot D_x(\sin\, x) \, dx $$ Now we can apply the formula of 'integration by parts' $$ \int \sin^2x\cdot\cos x \, dx = \int \sin^2x \cdot D_x(\sin\, x) \, dx = \sin^2x \cdot\sin\, x-\int D_x(\sin^2x)\cdot\sin\, x \, dx $$ By $D_x(\sin^2x)= 2\cdot \sin x\cdot D_x(\sin x)=2\cdot\sin x\cos x$ we have $$ \int \sin^2x\cdot\cos x \, dx = \sin^2x \cdot\sin\, x-\int D_x(\sin^2x)\cdot\sin\, x \, dx = \sin^3x-\int 2\cdot\sin x\cdot\cos x \cdot\sin\, x \, dx $$ and $$ \int \sin^2x\cdot\cos x \, dx = \sin^3x-2\int \sin^2\, x \cdot\cos x \, dx $$ Then we have $$ \int \sin^2\, x \cdot\cos x \, dx = \frac{1}{3}\sin^3 x $$

Answer 3

If you prefer making it complicated, the other way consists in linearising the integrand: \begin{align} \sin^2x\cos x&=\frac12(1-\cos 2x)\cos x=\frac12(\cos x-\cos 2x\cos x)\\ &=\frac12\Bigl(\cos x -\frac12\bigl(\cos(2x+x)+\cos(2x-x)\bigl)\Bigr)\\ &=\frac14(\cos x-\cos 3x). \end{align}

Answer 4

Use $2\sin(x)\cos(x)=\sin(2x)$:

$$I=\int\sin^2(x)\cos(x)\,dx=\frac{1}{2}\int\sin(x)\big(\frac{1}{2}(2\sin(x)\cos(x))\big)\,dx=\frac{1}{2}\int\sin(x)\sin(2x)\,dx$$

Two iterations of IBP yields: $$I=-\frac{1}{2}\sin(x)\cos(2x)+\frac{1}{2}\bigg[\frac{1}{2}\cos(x)\sin(2x)+\frac{1}{2}I\bigg]$$

Which becomes: $$I=\frac{4}{3}\Big(\frac{1}{4}\cos(x)\sin(2x)-\frac{1}{2}\sin(x)\cos(2x)\Big)+C$$

Answer 5

$$\int \sin^2x\,\cos\ x \, dx =\int \frac13(\sin^3x)'\,\cos\ x \, dx =\frac13\sin^3x+c$$