Why is substitution valid in Intergals? So i am a little confused why the u-substitution is valid. I do not have a problem when the integral is in the form $\int f(g(x))g'(x)dx=F(g(x))$ clearly then we can do $u=g(x)$ find $F(u)$ and do $F(g(x))$. However, things are not always so simple. Take the function $\int{\sqrt{1+x^2}x^5}$ the textbook uses the substitution $u=1+x^2$, however, quite clearly we no longer have the form $\int f(g(x))g'(x)dx$ or at least it is not as obvious. So basically it is a shoot in the dark? We HOPE that the substitution will actually give us that form?
Thanks.
 A: Actually you have that form in this case too, it is a little hidden. You can write
$$\int x^5\sqrt{1+x^2}dx=\int xx^4\sqrt{1+x^2}dx=\int\frac{2}{2}xx^4\sqrt{1+x^2}dx=\frac{1}{2}\int2xx^4\sqrt{1+x^2}dx.$$
Now you can notice that if you derive $1+x^2$ you get $2x$. So you try to make your integral easier by letting $1+x^2=y$, so $2xdx=dy$. 
Notice that $x^2=y-1$ so $x^4=(y-1)^2$. Hence
$$\frac{1}{2}\int2xx^4\sqrt{1+x^2}dx=\frac{1}{2}\int \sqrt{y}(y-1)^{2}dy$$
You can easily integrate it now by expanding the square and integrating with the power rule.
However is not always like this, but it can be useful to substitute even if the integrand is not like $f\left(g(x)\right)g'(x)$; for example
$$\int \sqrt{1-x^2}dx$$
If you substitute $x=\sin y$ you don't have the differential, but if you go on you get $dx=\cos y \ dy$ and so
$$\int\sqrt{1-x^2}dx=\int\sqrt{1-\left(\sin y\right)^{2}}\cos y \ dy=\int\left|\cos y\right|\cos y \ dy.$$
Which is pretty easy to integrate if you use the identity $\left(\cos y\right)^{2}=\frac{1+\cos (2x)}{2}.$
It's all about practice, experience, practice, knowing where you want to go and practice (did I mentioned practice?). 
