For example, $\int\sin^3x$ is turned into $\int sinxsin^2x$ then a substitution for $\sin^2x$ is made. What I would have done was $\int(sinx)^3$ and integrate via recognition: $\int (ax+b)^n dx$ =$\frac{(ax+b)^{n+1}}{a(n+1)} + C$ . However this will give a different answer from the correct one. Why is that? Why can't this reverse chain rule work for trigonometric functions?
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$\begingroup$ How does $\sin x$ correspond to $ax+b$? $\endgroup$ – Arthur Dec 28 '16 at 18:17
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1$\begingroup$ Because sine is not a linear function, so your rule for the power of a linear function won't apply. If you don't want to make a substitution, one recourse is to switch to complex exponentials. $\endgroup$ – J. M. isn't a mathematician Dec 28 '16 at 18:17
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1$\begingroup$ You would do as well to use the identity $\sin^2x=1-\cos^2x$ - no substitution required $\endgroup$ – Mark Bennet Dec 28 '16 at 18:25
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1$\begingroup$ You're also leaving out the "$dx$" in your notation which is absolutely critical for any substitution $\endgroup$ – MPW Dec 28 '16 at 18:34
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$\begingroup$ Apparently another person led astray by the notation $\sin x$ instead of $\sin(x)$. $\endgroup$ – Robert Israel Dec 28 '16 at 18:37
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This is basically boils down to the fact that $$ (x^n)' = n \, x^{n-1} $$ but $$ (\sin^n x)' = n \, \sin^{n-1} x \, \cos x. $$ You can see how these are not 'analogous' so you can't generalize the integral formula for $(a \, x + b)^n$
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Clearly this is not the case. For example, by your logic, we have
$$\int(x^2)^2\ dx=\frac13(x^2)^3+C=\frac13x^6+C$$
But it should be obvious that
$$\int(x^2)^2\ dx=\int x^4\ dx=\frac15x^5+C$$
If you can't do this with $x^2$, I don't see any reason why such logic would work on $\sin x$, which is far more complicated.
Particularly, you should differentiate your result to check if it is right, and don't forget to apply chain rule. For example,
$$\frac d{dx}\frac{\sin^4x}4=\cos x\sin^3x\ne\sin^3x$$
To do backwards chain rule, you needed a $\cos x$ out in the front, which there is none.
answered Dec 28 '16 at 18:18
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When trigonometric function have some power i.e non-linear functions. You can't do it simply. You have to use substitution or integration by parts.
In your case we have,
$\int\sin^3x$
=$\int sinxsin^2x$
On substitution,
=$\int sinx (1- cos^2x)$
Now put cos x = t then solve.
answered Dec 28 '16 at 18:29
