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This question already has an answer here:

So I've been trying to compute $$\int\sin^4(x)\mathrm{d}x$$ and everywhere they use the reduction formula which we haven't learned yet so I've been wondering if theres another way to do it? Thanks in advance.

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marked as duplicate by Hans Lundmark, Xander Henderson, Namaste calculus Mar 27 '18 at 21:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ are you familiar with integration by parts? $\endgroup$ – Arian Mar 27 '18 at 14:14
  • $\begingroup$ Unless the curriculum is very, very different in other parts of the world, the "reduction formula" is a basic trigonometric identity which you should have learned in a precalculus class long before ever enrolling in calculus. $\endgroup$ – Xander Henderson Mar 27 '18 at 21:34
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    $\begingroup$ @Sebastiano Why? What possible purpose could that serve? $\endgroup$ – Xander Henderson Mar 27 '18 at 21:34
  • $\begingroup$ Currently, it is not clear what "the reduction formula" in your question refers to. Please edit your question to be more precise about what it is. $\endgroup$ – user21820 Mar 28 '18 at 2:45
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Performing integration by parts,

$\begin{align} \int_0^x\sin^2 t\,dt&=\Big[-\cos t\sin t\Big]_0^x+\int_0^x\cos^2 t\,dt\\ &=-\cos x\sin x+\int_0^x(1-\sin^2 t)\,dt\\ &=-\cos x\sin x+\int_0^x 1\,dt-\int_0^x \sin^2 t\,dt\\ &=-\cos x\sin x+x-\int_0^x \sin^2 t\,dt\\ \end{align}$

Therefore,

$\displaystyle \int_0^x \sin^2 t\,dt=-\frac{1}{2}\cos x\sin x+\frac{1}{2}x$

$\begin{align} \int_0^x\sin^4 t\,dt&=\int_0^x(1-\cos^2)\sin^2 t\,dt \\ &=\int_0^x\sin^2 t\,dt-\int_0^x \cos^2 t\sin^2 t\,dt\\ &=-\frac{1}{2}\cos x\sin x+\frac{1}{2}x-\int_0^x \cos^2 t\sin^2 t\,dt\\ \end{align}$

Since, for $t$ real,

$\sin(2t)=2\sin t\cos t$

then,

$\begin{align} \int_0^x\sin^4 t\,dt&=-\frac{1}{2}\cos x\sin x+\frac{1}{2}x-\frac{1}{4}\int_0^x \sin^2(2t)\,dt\\ \end{align}$

In the latter integral perform the change of variable $y=2t$,

$\begin{align} \int_0^x\sin^4 t\,dt&=-\frac{1}{2}\cos x\sin x+\frac{1}{2}x-\frac{1}{8}\int_0^{2x} \sin^2(y)\,dy\\ &=-\frac{1}{2}\cos x\sin x+\frac{1}{2}x-\frac{1}{8}\left(-\frac{1}{2}\cos (2x)\sin(2x)+\frac{1}{2}\times 2x\right)\\ &=-\frac{1}{4}\sin(2x)+\frac{1}{2}x+\frac{1}{32}\sin(4x)-\frac{1}{8}x\\ &=-\frac{1}{4}\sin(2x)+\frac{3}{8}x+\frac{1}{32}\sin(4x)\\ \end{align}$

Therefore,

$\displaystyle \boxed{\int \sin^4 x\,dx=\frac{3}{8}x+\frac{1}{32}\sin(4x)-\frac{1}{4}\sin(2x)+C}$

($C$ a real constant)

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  • $\begingroup$ This appears to be the only answer that actually does what the questioner asks and performs the computation without using a power reduction formula or using other advanced techniques (such as complex integration) which are likely unfamiliar to the OP. Congratulations. $\endgroup$ – Xander Henderson Mar 27 '18 at 21:37
  • $\begingroup$ Thank you kindly <3 $\endgroup$ – Nicole Mar 28 '18 at 7:48
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Use this $$\sin^2x = \frac{1-\cos (2x)}2 \implies \sin^4x = \left( \frac{1-\cos 2x}2 \right)^2=\frac{1+\cos^2 (2x) -2 \cos 2x}{4}$$

And then,

$$\cos^2 (2x)=\frac{1+\cos (4x) }{2}$$

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An alternative way might be to use the fact that $$\sin(x)=\frac{1}{2}i \left(e^{-ix}-e^{ix}\right)$$ $$\sin^4(x)=\frac{1}{16} \left(e^{-ix}-e^{ix}\right)^4$$ And you just need to expand and integrate it.

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  • $\begingroup$ In my opinion, this is not just an alternative way, but the definitive way to do it, since it immediately explains all the coefficients of the terms. $\endgroup$ – user21820 Mar 28 '18 at 2:42

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