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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.


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

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  • $\begingroup$ 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
  • 2
    $\begingroup$ @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

Performing integration by parts,

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


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

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

Since, for $t$ real,

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


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

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

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


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

($C$ a real constant)

  • $\begingroup$ 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

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,

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


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.

  • $\begingroup$ 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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