Math, asked by anuradasrinivasan, 1 month ago

integral of(x³+3x²+6x+6)cos2x dx using bernoulli's formula​

Answers

Answered by MaheswariS
2

\textbf{Given:}

\mathsf{\displaystyle\int\,(x^3+3x^2+6x+6)cos2x\;dx}

\textbf{To evaluate:}

\mathsf{\displaystyle\int\,(x^3+3x^2+6x+6)cos2x\;dx}

\textbf{Solution:}

\underline{\textsf{Bernoulli's formula:}}

\mathsf{\displaystyle\int\,u\,dv=uv-u'v_1+u''v_2-u'''v_3+\;.\;.\;.\;.\;.}

\mathsf{Take,}

\mathsf{u=x^3+3x^2+6x+6}

\mathsf{u'=3x^2+6x+6}

\mathsf{u''=6x+6}

\mathsf{u'''=6}

\mathsf{and}

\mathsf{dv=cos2x\,dx}

\mathsf{\displystyle\int\,dv=\int\,cos2x\,dx}

\mathsf{v=\dfrac{sin2x}{2}}

\mathsf{v_1=\dfrac{-cos2x}{4}}

\mathsf{v_2=\dfrac{-sin2x}{8}}

\mathsf{v_3=\dfrac{cos2x}{16}}

\mathsf{By\;Bernoulli's\;formula,}

\mathsf{\displaystyle\int\,(x^3+3x^2+6x+6)cos2x\;dx}

\mathsf{=(x^3+3x^2+6x+6)\dfrac{sin2x}{2}-(3x^2+6x+6)\left(\dfrac{-cos2x}{4}\right)+(6x+6)\left(\dfrac{-sin2x}{8}\right)-6\left(\dfrac{cos2x}{16}\right)+C}

\mathsf{=(x^3+3x^2+6x+6)\dfrac{sin2x}{2}+(3x^2+6x+6)\left(\dfrac{cos2x}{4}\right)-6(x+1)\left(\dfrac{sin2x}{8}\right)-3\left(\dfrac{cos2x}{8}\right)+C}

\mathsf{=(x^3+3x^2+6x+6)\dfrac{sin2x}{2}+(3x^2+6x+6)\left(\dfrac{cos2x}{4}\right)-3(x+1)\left(\dfrac{sin2x}{4}\right)-3\left(\dfrac{cos2x}{8}\right)+C}

\textbf{Find more:}

Limit 0 to Pi /2 integration xsinx

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