$I=\int \frac{\cos^3(x)}{\sqrt{\sin^7(x)}}\,dx$

Question

$$I=\int \frac{\cos^3(x)}{\sqrt{\sin^7(x)}}\,dx$$ 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?

Answer 1

Hint: $$I= \int\frac{(1-\sin^2 x)\cos x}{\sqrt{\sin^7x}}dx$$ Substitution: $\sin x=t$.

Answer 2

$$ \text{Hint:}\quad \int \frac{\cos^2(x)}{\sqrt{\sin^7(x)}} \underbrace{\Big( \cos x \,dx\Big)} $$

If you don't know how to construe this as a hint, then your understanding of integration by substitution isn't all there.

Answer 3

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 $$\int\frac{x}{\sqrt{1-x^2}}dx=\int\frac{x\cdot\frac{dx}{1}}{\sqrt{1-x^2}}=\int\frac{x\cdot\frac{dx}{1}}{\sqrt{1-x^2}}\cdot\frac{\frac{d(1-x^2)}{dx}}{\frac{d(1-x^2)}{dx}}=\int\frac{x}{\sqrt{1-x^2}}\cdot\frac{d(1-x^2)}{-2x}$$ $$=\int\frac{1}{-2\sqrt{1-x^2}}d(1-x^2)=\int\frac{1}{-2\sqrt{u}}du=-\sqrt{u}+C=-\sqrt{1-x^2}+C$$

Answer 4

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 $$ \int \frac{1-u^2}{\sqrt{u^7}}du = \int \frac{1}{\sqrt{u^7}}-\frac{u^2}{\sqrt{u^7}}du $$ $$= \int u^{-\frac72}du-\int u^{-\frac32}du $$ $$= \frac{u^{-\frac72+1}}{-\frac72+1}- \frac{u^{-\frac32+1}}{-\frac32+1}+C $$ $$= \frac{u^{-\frac52}}{-\frac52}- \frac{u^{-\frac12}}{-\frac12} +C $$ $$= 2u^{-\frac12} -\frac{2}{5}u^{-\frac52}+C$$ $$= \frac{2}{\sqrt{u}}-\frac{2}{5\sqrt{u^{5}}}+C$$ $$= \frac25\left(\frac{5\sqrt{u^5}-\sqrt{u}}{\sqrt{u}\sqrt{u^{5}}}\right) +C$$ $$= \frac25\left(\frac{5u^2-1}{\sqrt{u^{5}}}\right) +C$$ $$= \frac25\left(\frac{5\sin^2(x)-1}{\sqrt{\sin^{5}(x)}}\right) +C$$