Why can't I integrate trigonometric functions without making a substitution? 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?
 A: 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 $(a \, x + b)^n$
A: 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.
A: 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.
