$\int_{-\pi / 4}^{\pi / 4} x^3 \sin ^4(x) d x=$

Question

Aicharya · Accepted Answer

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

\(\int_{-\pi / 4}^{\pi / 4} x^3 \sin ^4(x) d x=\)

Solution