$I=\int \frac{\cos^3(x)}{\sqrt{\sin^7(x)}}\,dx$ $$I=\int \frac{\cos^3(x)}{\sqrt{\sin^7(x)}}\,dx$$
I tried to write it as 
$$I=\int \sqrt{\frac{\cos^6(x)}{\sin^7(x)}}\,dx$$
And $$I=\int \sqrt{\frac{1}{\tan^6(x)\sin(x)}}\,dx$$ but it seems to go nowhere , how can I manipulate it so that its becomes solvable?
 A: Here are the steps
$$\int \frac{\cos^3 x}{\sqrt{\sin^7 x}}dx= \int \frac{(1-\sin^2 x)(\cos x)}{\sqrt{\sin^7 x}}dx $$
Let $u=\sin x$, then $du=\cos x\ dx$. So now
$$ \int \frac{1-u^2}{\sqrt{u^7}}du = \int \frac{1}{\sqrt{u^7}}-\frac{u^2}{\sqrt{u^7}}du $$
$$= \int u^{-\frac72}du-\int u^{-\frac32}du $$
$$= \frac{u^{-\frac72+1}}{-\frac72+1}- \frac{u^{-\frac32+1}}{-\frac32+1}+C $$
$$= \frac{u^{-\frac52}}{-\frac52}- \frac{u^{-\frac12}}{-\frac12} +C $$
$$= 2u^{-\frac12} -\frac{2}{5}u^{-\frac52}+C$$
$$= \frac{2}{\sqrt{u}}-\frac{2}{5\sqrt{u^{5}}}+C$$
$$= \frac25\left(\frac{5\sqrt{u^5}-\sqrt{u}}{\sqrt{u}\sqrt{u^{5}}}\right) +C$$
$$= \frac25\left(\frac{5u^2-1}{\sqrt{u^{5}}}\right) +C$$
$$= \frac25\left(\frac{5\sin^2(x)-1}{\sqrt{\sin^{5}(x)}}\right) +C$$
A: Hint:
$$I= \int\frac{(1-\sin^2 x)\cos x}{\sqrt{\sin^7x}}dx$$
Substitution: $\sin x=t$.
A: $$
\text{Hint:}\quad \int \frac{\cos^2(x)}{\sqrt{\sin^7(x)}} \underbrace{\Big( \cos x \,dx\Big)}
$$
If you don't know how to construe this as a hint, then your understanding of integration by substitution isn't all there.
A: I'm going to suggest that we change the way that we present substitution. One problem with understanding substitution is that it is taught as a procedure, that somehow magically allows you to come up with the right answer. For example, in this problem we would define
$$u=\sin x$$
differentiate, and solve for $du$
$$\frac{du}{dx} = \cos x$$
$$du = \cos x dx$$
then we try to find that loose $\cos x$ and hope that everything works out for the best.
At some point, I started doing this:
$$\int \frac{\cos^3x}{\sqrt{\sin^7 x}}dx = \int \frac{\cos^3x}{\sqrt{\sin^7 x}}dx\cdot\frac{\frac{du}{dx}}{\color{green}{\frac{du}{dx}}}$$
I interpret the numerator ($du/dx$) as a ratio of infinitesimal  changes in $\sin x$ (i.e. $du$) and  $x$ (i.e. $dx$), making a mental note that $u=\sin x$. This allows me to just cancel out the $dx$s. The denominator ($du/dx$) is interpreted as a derivative, which evaluates to $\frac{du}{dx}=\frac{d\sin x}{dx}=\cos x$.
My next step would be as follows:
$$\int \frac{\cos^3x}{\sqrt{\sin^7 x}}\cdot\frac{\frac{du\cdot dx}{dx}}{\color{green}{\cos x}}=\int \frac{\cos^2x}{\sqrt{\sin^7 x}}du=\int \frac{1-\sin^2x}{\sqrt{\sin^7 x}}du=\int \frac{1-u^2}{\sqrt{u^7}}du$$
When we present substitution this way, I think that it is much more clear that we divide by the derivative of the substitution, than it is when we use the magical procedural approach.
Consider another example
$$\int\frac{x}{\sqrt{1-x^2}}dx$$
notice that if we divide by the derivative of $1-x^2$ (i.e. $-2x$), that the loose $x$ in the numerator cancels out, so $u=1-x^2$ would make a good choice for a substitution.
Integrating using this approach would look like this
$$\int\frac{x}{\sqrt{1-x^2}}dx=\int\frac{x\cdot\frac{dx}{1}}{\sqrt{1-x^2}}=\int\frac{x\cdot\frac{dx}{1}}{\sqrt{1-x^2}}\cdot\frac{\frac{d(1-x^2)}{dx}}{\frac{d(1-x^2)}{dx}}=\int\frac{x}{\sqrt{1-x^2}}\cdot\frac{d(1-x^2)}{-2x}$$
$$=\int\frac{1}{-2\sqrt{1-x^2}}d(1-x^2)=\int\frac{1}{-2\sqrt{u}}du=-\sqrt{u}+C=-\sqrt{1-x^2}+C$$
