Finding $\int^{\pi}_{0}x^4\sin^4(x)dx$

Question

Finding $\displaystyle \int^{\pi}_{0}x^4\sin^4(x)dx$

Try: $$I= \int^{\pi}_{0}x^4\sin^4(x)dx= \int^{\pi}_{0}(\pi-x)^4\sin^4(x)dx$$

$$I=\int^{\pi}_{0}\bigg(\pi^4-4\pi^3x+6\pi^2x^2-4\pi x+x^4\bigg)\sin^4 xdx$$

Could some help me to solve it, Thanks

Answer 1

Since $$\sin^4 x = \frac{1}{8} \cos (4x) - \frac{1}{2} \cos (2x) + \frac{3}{8},$$ the integral can be rewritten as $$\int_0^\pi x^4 \sin^4 x \, dx =\frac{1}{8} \int_0^\pi x^4 \cos (4x) \, dx - \frac{1}{2} \int_0^\pi x^4 \cos (2x) + \frac{3}{8} \int_0^\pi x^4 dx. \tag1$$

Now consider $$I(a) = \int_0^\pi \cos (ax) \, dx, \quad a \geqslant 0.$$ Differentiating with respect to the parameter $a$ four times gives $$I^{(4)} (a) = \int_0^\pi x^4 \cos (ax) \, dx. \tag2$$ So in terms of (2) the integral appearing in (1) can be written as $$\int_0^\pi x^4 \sin^4 x \, dx = \frac{1}{8} I^{(4)}(4) - \frac{1}{2} I^{(4)}(2) + \frac{3}{8} I^{(4)}(0).$$

Now, as $$I(a) = \int_0^\pi \cos (ax) \, dx = \frac{\sin (a \pi)}{a},$$ we have $$I^{(4)}(a) = \frac{1}{a^5} \big{[} 4\pi a (\pi^2 a^2 - 6) \cos (\pi a) + (\pi^4 a^4 - 12 \pi^2 a^2 + 24) \sin (\pi a) \Big{]},$$ and we see that $$I^{(4)}(4) = \frac{\pi^3}{4} - \frac{3\pi}{32}, \quad I^{(4)}(2) = \pi^3 - \frac{3\pi}{2}, \quad I^{(4)}(0) = \frac{\pi^5}{5},$$ yielding $$\int_0^\pi x^4 \sin^4 x \, dx = \frac{1}{8} \left (\frac{\pi^3}{4} - \frac{3\pi}{32} \right ) - \frac{1}{2} \left (\pi^3 - \frac{3\pi}{2} \right ) + \frac{3 \pi^5}{40} = \frac{3}{40} \pi^5 - \frac{15}{32} \pi^3 + \frac{189}{256} \pi.$$

Answer 2

Hint

Try to reduce the degree of the sine function

$$I=\int^{\pi}_{0}x^4\sin^4(x)dx=\int^{\pi}_{0}x^4\sin^2(x)(1-\cos^2(x))dx=\int^{\pi}_{0}x^4(\sin^2(x)-\frac 1 4 \sin^2(2x))dx=....$$

Use again $\sin^2(x)=\frac {1-\cos(2x)} 2$ to reduce again

After that integrate by part...

Answer 3

Focusing on the antiderivative, let use use $$\sin^4(x)=\frac{1}{8} (\cos (4 x)-4 \cos (2 x)+3)$$ $$8I =\int x^4 \cos(4x)\,dx-4\int x^4 \cos(2x)\,dx+3x$$ $$8I=\frac1 {1024}\int (4x)^4\cos(4x)\,d(4x)-\frac 18\int(2x)^4\cos(2x)\,d(2x)+3x$$

Answer 4

Hint: Use symmetry property and let $u=\dfrac{\pi}{2}-x$ \begin{align} \int^{\pi}_{0}x^4\sin^4(x)dx &= \int^{-\frac{\pi}{2}}_{\frac{\pi}{2}}\left(\frac{\pi}{2}-u\right)^4\cos^4(\frac{\pi}{2}-u)d(\frac{\pi}{2}-u) \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left((\frac{\pi}{2})^4-4(\frac{\pi}{2})^3u+6(\frac{\pi}{2})^2u^2-4(\frac{\pi}{2})^1u^3+u^4\right)\cos^4u\,\mathrm{du}\\ &= (\frac{\pi}{2})^4\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^4u\,\mathrm{du}+6(\frac{\pi}{2})^2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}u^2\cos^4u\,\mathrm{du}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}u^4\cos^4u\,\mathrm{du} \end{align} then use $\cos^4u=\dfrac18\left(3+2\cos2x+\cos4x\right)$.