Question

Solve the integral:$\int x^2\cos xdx$.

Short answer

The required answer is$x^2\sin x+2x\cos x-2\sin x+c$.

Step-by-step solution

Step 1. Given information. 

We have given integral is$\int x^2\cos xdx$.

Step 2. Solve the integration by parts . 

We have, $u=x^{2}dv=\cos xdx$

$$\begin{gathered} u=x^2 \\ du=2xdx \end{gathered}$$

and

$$\begin{gathered} dv=\cos xdx \\ v=\int \cos xdx \\ v=\sin x \end{gathered}$$

The formula of integration by parts is $\int udv=uv-\int vdu$.

$$\begin{gathered} \int x^2\cos xdx \\ =x^2\sin x-\int 2x\sin xdx \\ =x^2\sin x-2\int x\sin xdx \end{gathered}$$

Step 3. Integration by parts.

We have,

$$\begin{gathered} u=x,dv=\sin xdx \\ u=x \\ du=dx \end{gathered}$$

and

$$\begin{gathered} dv=\sin xdx \\ v=\int \sin xdx \\ v=-\cos x \end{gathered}$$

So,

$$\begin{gathered} x^2\sin x-2\int x\sin xdx \\ =x^2\sin x-2\left(\right(x(-\cos x))-\int (-\cos x\left)dx\right) \\ =x^2\sin x-2\left(\right(x(-\cos x))+\int \cos xdx) \\ =x^2\sin x+2x\cos x-2\sin x+c \end{gathered}$$