Substitution in $\int \cos^3(x)$

Question

I have an problem with the substitution in

$$\int \cos^3(x)dx$$

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Ive seen some answers where the integration is done by substituting

$$\int \cos^3(x) = \int \cos(x)(1-\sin^{2}(x))$$

$u = \sin(x)$

I understand how we get the function in that form (trigonometric identity $\cos^2(x) + \sin^2(x) = 1 \implies \cos^2(x) = 1 - \sin^2(x)$, but I dont understand how the substitution can be done with $u = \sin(x)$, since the function needs to be in form $\int f(g(x))g'(x)$ before we can do the substitution to $u = g(x)$? Could you help me see what the $g(x)$ and $f(x)$ is here, thanks!

Answer 1

$$g(x) = \sin (x) \, $$

$$f(x) = 1-x^2$$

We indeed have our integration in the form of $\int g'(x)f(g(x)) \, dx$.

Answer 2

Note that, if $f(x)=1-x^2$, then$$\cos(x)\bigl(1-\sin^2(x)\bigr)=f\bigl(\sin(x)\bigr)\cos(x)=f\bigl(\sin(x)\bigr)\sin'(x).$$

Answer 3

$\displaystyle\int \cos x(1-\sin^2 x)\,dx$

$=\displaystyle\int\cos x\,dx-\displaystyle\int\sin^2 x\cos x\,dx$

$=\displaystyle\int d(\sin x)\,dx-\displaystyle\int\sin^2x\cdot d(\sin x)\,dx$

$=\sin x-\dfrac{\sin^3x}{3}+C$

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There is another elegant way to solve such integrals which just require to know that $\left(a+\dfrac1a\right)^3=\left(a^3+\dfrac{1}{a^3}\right)+3\left(a+\dfrac1a\right)$ and the identities: $\cos x=\dfrac{e^{ix}+e^{-ix}}{2},\sin x=\dfrac{e^{ix}-e^{-ix}}{2i}.$

$\therefore\displaystyle\int \cos^3x\,dx$

$=\dfrac18\displaystyle\int(e^{ix}+e^{-ix})^3\,dx$

$=\dfrac18\displaystyle\int e^{3ix}+e^{-3ix}+3\left(e^{ix}+e^{-ix}\right)\,dx$

$=\dfrac{e^{3ix}-e^{-3ix}}{24i}+3\dfrac{e^{ix}-e^{-ix}}{8i}$

$=\dfrac{\sin 3x}{12}+\dfrac{3\sin x}{4}+C$