How do I evaluate this definite integral? $$\int_{0}^{\frac{\pi}{12}}{\sin^4x \, \cos^4x\, \operatorname{d}\!x}$$ I know this is a trig. function.
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2Use $\frac{1}{2}\sin(2x)=\cos(x)\sin(x)$. – Meow Mar 07 '13 at 17:15
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From here, we have $$\int_0^{\pi/2} \sin^{2n}(x) dx = \dfrac{2n-1}{2n} \cdot \dfrac{2n-3}{2n-2} \cdots \dfrac34 \cdot \dfrac12 \cdot\dfrac{\pi}2$$ Hence, $$\int_0^{\pi/2} \sin^4(x) \cos^4(x) dx = \dfrac1{16} \int_0^{\pi/2} \sin^4(2x) dx = \dfrac1{32} \int_0^{\pi} \sin^4(t) dt = \dfrac2{32} \int_0^{\pi/2} \sin^4(t) dt$$ Hence, the answer is $$\dfrac1{16} \cdot \dfrac34 \cdot \dfrac12 \cdot \dfrac{\pi}2 = \dfrac{3 \pi}{256}$$
If the integral is from $0$ to $\pi/12$, then, we get that \begin{align} \int_0^{\pi/{12}} \sin^4(x) \cos^4(x) dx & = \dfrac1{16} \int_0^{\pi/12} \sin^4(2x) dx = \dfrac1{32} \int_0^{\pi/6} \sin^4(t) dt\\ & = \dfrac1{32} \int_0^{\pi/6} \dfrac{(1-\cos(2t))^2}{4} dt\\ & = \dfrac1{128} \int_0^{\pi/6} (1+\cos^2(2t) - 2 \cos(2t)) dt\\ & = \dfrac1{128} \int_0^{\pi/6} \left(1+\dfrac{1+\cos(4t)}2 - 2 \cos(2t) \right) dt\\ & = \dfrac1{128} \int_0^{\pi/6} \left(\dfrac32+\dfrac{\cos(4t)}2 - 2 \cos(2t) \right) dt\\ & = \dfrac1{128} \left(\dfrac{\pi}4 + \dfrac{\sin(4 \pi/6)}{8}-\sin(2 \pi/6)\right)\\ & = \dfrac1{128} \left(\dfrac{\pi}4 + \dfrac{\sqrt3}{16} - \dfrac{\sqrt3}2\right)\\ & = \dfrac1{128} \left(\dfrac{\pi}4 - \dfrac{7\sqrt3}{16} \right) = \dfrac{4 \pi - 7 \sqrt3}{2048} \end{align}
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We have $$\sin^4x\cos^4x=\frac{1}{16}\sin^4(2x)=\frac{1}{16}(\frac{e^{2ix}-e^{-2ix}}{2i})^4=\frac{1}{16^2}(2\cos(8x)-8\cos(4x)+6), $$ so we integrate and we find $$\int_{0}^{\frac{\pi}{12}}{\sin^4x \, \cos^4x\, \operatorname{d}\!x}=\frac{1}{512}π-\frac{7}{2048}\sqrt{3}.$$
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You used the wrong integration limits, but I like the approach. – Nathaniel Bubis Mar 07 '13 at 17:51
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