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\(\int_{-\pi / 4}^{\pi / 4} x^3 \sin ^4(x) d x=\)
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\(I=\int_{-\pi / 4}^{\pi / 4} x^3 \cdot \sin ^4 x d x\)
As, \(\int_{-a}^a f(x) d x=0\), when
\(f(-x)=-f(x),\)
We have,
\(\begin{gathered}
f(-x)=(-x)^3 \sin ^4(-x) \\
=-\left(x^3 \cdot \sin ^4 x\right)=-f(x) \\
\therefore \quad \int_{-\pi / 4}^{\pi / 4} x^3 \cdot \sin ^4 x \cdot d x=0 .
\end{gathered}\)
As, \(\int_{-a}^a f(x) d x=0\), when
\(f(-x)=-f(x),\)
We have,
\(\begin{gathered}
f(-x)=(-x)^3 \sin ^4(-x) \\
=-\left(x^3 \cdot \sin ^4 x\right)=-f(x) \\
\therefore \quad \int_{-\pi / 4}^{\pi / 4} x^3 \cdot \sin ^4 x \cdot d x=0 .
\end{gathered}\)
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