Question

Evaluating the integral

\(\int {({x^2} + 2x)\cos xdx} \)

Short answer

• We should use the substitution method to solve the integral
• We will apply integration by parts multiple times. We should use

\(i = uv\int {vdu} \)This is to solve the integral

Step-by-step solution

-Given Data

Given- Let,\(i = \int {({x^2} + 2x)\cos xdx} \)

Let u=\({x^2} + 2x\) \(dv = {\rm{ }}cosx{\rm{ }}dx\]

\(du = \left( {2x + 2} \right){\rm{ }}dx\;\;\;\;\;\;v = {\rm{ }}sin{\rm{ }}x\)\(du = \left( {2x + 2} \right){\rm{ }}dx\;\;\;\;\;\;v = {\rm{ }}sin{\rm{ }}x\)

\(i = ({x^2} + 2x)\sin x - \int {(2x + 2)\sin xdx} \)

This is simpler if we write it,

\(i = ({x^2} + 2x)\sin x + \int {(2x = 2)( - \sin x)dx} \)

Reassigning value

Let \(\begin{array}{*{20}{l}}{u{\rm{ }} = {\rm{ }}2x + 2\;\;\;\;\;\;\;\;\;\;\;dv = {\rm{ }} - sinx{\rm{ }}dx}\\{du = 2{\rm{ }}dx\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;v = {\rm{ }}cos{\rm{ }}x}\end{array}\]

\(\begin{array}{l}i = ({x^2} + 2x)\sin x + uv - \int {vdu} \\i = ({x^2} + 2x)\sin x + (2x + 2)\cos x - \int {2\cos xdx} \\i = ({x^2} + 2x)\sin x + (2x + 2)\cos x - 2\sin x\\i = ({x^2} + 2x - 2)\sin x + (2x + 2)\cos x + c\end{array}\)

Hence, c is constant

Therefore, the integral value of \(\int {({x^2} + 2x)\cos xdx} \)

\(({x^2} + 2x - 2)\sin x + (2x + 2)\cos x + c\)