Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
Use $\frac{1}{2}\sin(2x)=\cos(x)\sin(x)$. –  Alyosha Mar 7 '13 at 17:15

2 Answers 2

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}

share|cite|improve this answer

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 $$\int_{0}^{\frac{\pi}{12}}{\sin^4x \, \cos^4x\, \operatorname{d}\!x}=\frac{1}{512}π-\frac{7}{2048}\sqrt{3}.$$

share|cite|improve this answer
You used the wrong integration limits, but I like the approach. –  nbubis Mar 7 '13 at 17:51
@nbubis To err is human. –  user63181 Mar 7 '13 at 17:59
And to upvote after the answer is corrected is also. –  nbubis Mar 7 '13 at 18:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.