I need help finding out how to Find $\displaystyle \int \frac{cosx}{sin^2x} dx$.
How should I go about this?
I seriously doubt you have the correct solution, as your solution process is completely off...
If you let $\displaystyle \displaystyle \begin{align*} u = \sin{x} \end{align*} $ then $\displaystyle \displaystyle \begin{align*} du = \cos{x}\,dx \end{align*} $
So the integral becomes
$\displaystyle \displaystyle \begin{align*} \int{\frac{\cos{x}}{\sin^2{x}}\,dx} &= \int{\frac{1}{u^2}\,du} \\ &= \int{u^{-2}\,du} \end{align*} $
Not anything which has a $\displaystyle \displaystyle \begin{align*} \cos{x} \end{align*} $ or any of its variants in the integrand...
Hello Prove it,
i ended up with the same thing as you. Here is my reasoning.
$\displaystyle \int \frac{cosx}{sin^2x} dx$
u = sin(x) u'=cos(x)
I must change the function in terms of u, so I start with the denominator.
$\displaystyle \int \frac{cosx}{u^2} dx$
Now I will change my numerator cosx, in terms of u. Using the trigonometric Identity$\displaystyle sin^2x+cos^2x=1$
$\displaystyle cosx= \sqrt{1-sin^2x} $ $\displaystyle \Rightarrow \sqrt{1-u^2}$
so now I have ....
$\displaystyle \int \frac{\sqrt{1-u^2}}{u^2} dx$
Now I must change the constant multiplier of dx. I am going to use $\displaystyle \sqrt{1-u^2} dx$
Using $\displaystyle dx= \sqrt{1-u^2} $ & du or u'=cosx, I set them equal to eachother. NOTE: I must also change u' in terms of u
$\displaystyle \sqrt{1-u^2} dx =cosx du \Rightarrow dx =\frac{cosx du}{\sqrt{1-u^2} } \Rightarrow \frac{dx}{du} =\frac{cosx }{\sqrt{1-u^2} } $
So now I have dx in terms of u, $\displaystyle \frac{cosx }{\sqrt{1-u^2} } $ Notice that cosx still has an x. I must remember to keep everything in terms of u. So remembering back to the trigonometric identity, $\displaystyle cosx= \sqrt{1-sin^2x} $ I now have
$\displaystyle \frac{\sqrt{1-u^2} }{\sqrt{1-u^2} } $ = 1
I plug that back into my new integral
$\displaystyle \int \frac{1}{u^2} du$
$\displaystyle \int u^{-2} du$
Your reasoning, though possibly correct, is extremely convoluted and difficult to follow.
If you let $\displaystyle \displaystyle \begin{align*} u = \sin{x} \end{align*} $ then $\displaystyle \displaystyle \begin{align*} \frac{du}{dx} = \cos{x} \implies du = \cos{x}\,dx \end{align*} $
Can you see that $\displaystyle \displaystyle \begin{align*} \cos{x}\,dx \end{align*} $ appears in your integral? Just change it to $\displaystyle \displaystyle \begin{align*} du \end{align*} $...
Can I just point out a shorthand approach, using the formula:
$\displaystyle \int f'(x)[f(x)]^n}~dx=\frac{[f(x)]^{n+1}}{n+1}+C$? Obviously, working at developing your ability to use the u-substitution is highly beneficial, so I'm not discouraging you from using it, but this approach would undoubtedly save you time and energy. Here, $\displaystyle f(x)=\sin(x)$, $\displaystyle n=-2$ and $\displaystyle f'(x)=\cos(x)$
Edit: You could also use the fact that $\displaystyle \frac{d}{dx}\csc(x)=\csc(x)\cot(x)$ and simply change the sign appropriately.