# Thread: find integral w/ trig

1. ## find integral w/ trig

I need help finding out how to Find $\int \frac{cosx}{sin^2x} dx$.

2. ## Re: find integral w/ trig

Try $u = \sin x \implies \frac{du}{dx}= \dots$

3. ## Re: find integral w/ trig

$\int \frac{cosx}{u^2} dx$

How would I change the variable of the the numator so it is in term of u

u = sinx
it cant be this
$\int \frac{u-u + cosx}{u^2} dx$

4. ## Re: find integral w/ trig

ahh $cosx = \sqrt{ -u^2+1}$ via the trig identity, doh!

5. ## Re: find integral w/ trig

fantastic, I got the solution now. thanks!

6. ## Re: find integral w/ trig

I seriously doubt you have the correct solution, as your solution process is completely off...

If you let \displaystyle \begin{align*} u = \sin{x} \end{align*} then \displaystyle \begin{align*} du = \cos{x}\,dx \end{align*}

So the integral becomes

\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 \begin{align*} \cos{x} \end{align*} or any of its variants in the integrand...

7. ## Re: find integral w/ trig

Originally Posted by Prove It
I seriously doubt you have the correct solution, as your solution process is completely off...

If you let \displaystyle \begin{align*} u = \sin{x} \end{align*} then \displaystyle \begin{align*} du = \cos{x}\,dx \end{align*}

So the integral becomes

\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 \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.

$\int \frac{cosx}{sin^2x} dx$

u = sin(x) u'=cos(x)

I must change the function in terms of u, so I start with the denominator.

$\int \frac{cosx}{u^2} dx$

Now I will change my numerator cosx, in terms of u. Using the trigonometric Identity $sin^2x+cos^2x=1$

$cosx= \sqrt{1-sin^2x}$ $\Rightarrow \sqrt{1-u^2}$

so now I have ....

$\int \frac{\sqrt{1-u^2}}{u^2} dx$

Now I must change the constant multiplier of dx. I am going to use $\sqrt{1-u^2} dx$

Using $dx= \sqrt{1-u^2}$ & du or u'=cosx, I set them equal to eachother. NOTE: I must also change u' in terms of u

$\sqrt{1-u^2} dx =cosx du \Rightarrow dx =\frac{cosx du}{\sqrt{1-u^2} } \Rightarrow \frac{dx}{du} =\frac{cosx }{\sqrt{1-u^2} }$

So now I have dx in terms of u, $\frac{cosx }{\sqrt{1-u^2} }$ Notice that cosx still has an x. I must remember to keep everything in terms of u. So remembering back to the trigonometric identity, $cosx= \sqrt{1-sin^2x}$ I now have

$\frac{\sqrt{1-u^2} }{\sqrt{1-u^2} }$ = 1
I plug that back into my new integral

$\int \frac{1}{u^2} du$

$\int u^{-2} du$

8. ## Re: find integral w/ trig

Originally Posted by delgeezee
Hello Prove it,

i ended up with the same thing as you. Here is my reasoning.

$\int \frac{cosx}{sin^2x} dx$

u = sin(x) u'=cos(x)

I must change the function in terms of u, so I start with the denominator.

$\int \frac{cosx}{u^2} dx$

Now I will change my numerator cosx, in terms of u. Using the trigonometric Identity $sin^2x+cos^2x=1$

$cosx= \sqrt{1-sin^2x}$ $\Rightarrow \sqrt{1-u^2}$

so now I have ....

$\int \frac{\sqrt{1-u^2}}{u^2} dx$

Now I must change the constant multiplier of dx. I am going to use $\sqrt{1-u^2} dx$

Using $dx= \sqrt{1-u^2}$ & du or u'=cosx, I set them equal to eachother. NOTE: I must also change u' in terms of u

$\sqrt{1-u^2} dx =cosx du \Rightarrow dx =\frac{cosx du}{\sqrt{1-u^2} } \Rightarrow \frac{dx}{du} =\frac{cosx }{\sqrt{1-u^2} }$

So now I have dx in terms of u, $\frac{cosx }{\sqrt{1-u^2} }$ Notice that cosx still has an x. I must remember to keep everything in terms of u. So remembering back to the trigonometric identity, $cosx= \sqrt{1-sin^2x}$ I now have

$\frac{\sqrt{1-u^2} }{\sqrt{1-u^2} }$ = 1
I plug that back into my new integral

$\int \frac{1}{u^2} du$

$\int u^{-2} du$
Your reasoning, though possibly correct, is extremely convoluted and difficult to follow.

If you let \displaystyle \begin{align*} u = \sin{x} \end{align*} then \displaystyle \begin{align*} \frac{du}{dx} = \cos{x} \implies du = \cos{x}\,dx \end{align*}

Can you see that \displaystyle \begin{align*} \cos{x}\,dx \end{align*} appears in your integral? Just change it to \displaystyle \begin{align*} du \end{align*}...

9. ## Re: find integral w/ trig

Originally Posted by Prove It
Your reasoning, though possibly correct, is extremely convoluted and difficult to follow.

If you let \displaystyle \begin{align*} u = \sin{x} \end{align*} then \displaystyle \begin{align*} \frac{du}{dx} = \cos{x} \implies du = \cos{x}\,dx \end{align*}

Can you see that \displaystyle \begin{align*} \cos{x}\,dx \end{align*} appears in your integral? Just change it to \displaystyle \begin{align*} du \end{align*}...

I agree. The more I do, the easier it will be for me to identify patterns.

10. ## Re: find integral w/ trig

Can I just point out a shorthand approach, using the formula:

$\int f'(x)[f(x)]^n}~dx=\frac{[f(x)]^{n+1}}{n+1}+C$? 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, $f(x)=\sin(x)$, $n=-2$ and $f'(x)=\cos(x)$

Edit: You could also use the fact that $\frac{d}{dx}\csc(x)=\csc(x)\cot(x)$ and simply change the sign appropriately.