# Thread: indefinite integral of 4cos+sinx-cosx/ sin^2x

1. ## indefinite integral of 4cos+sinx-cosx/ sin^2x

4cos+sinx-cosx/ sin^2x
How do I start? I don't think I can use u substitution:/
I tried canceling the sinx bit I'm still left with sinx at the bottom.

2. ## Re: indefinite integral of 4cos+sinx-cosx/ sin^2x

Hello Dream,

We have the following:
$\displaystyle \int \frac{4cosx+sinx-cosx}{sin^2x}dx \implies \int \frac{3cosx+sinx}{sin^2x}dx$

$\displaystyle =3\int \frac{cosx}{sin^2x}dx+ \int \frac{1}{sinx}dx$

Can you finish it?

[hint] For the first substitution, let u = sinx , for the second write $\displaystyle \frac{1}{sinx}=cscx$ [/hint]

3. ## Re: indefinite integral of 4cos+sinx-cosx/ sin^2x

Originally Posted by dreamd7358
4cos+sinx-cosx/ sin^2x
How do I start? I don't think I can use u substitution:/
I tried canceling the sinx bit I'm still left with sinx at the bottom.
well, what if the integrand is as follows?

$\displaystyle \int 4\cos{x} + \sin{x} - \frac{\cos{x}}{\sin^2{x}} \, dx =$

$\displaystyle \int 4\cos{x} + \sin{x} - \csc{x}\cot{x} \, dx = 4\sin{x} - \cos{x} + \csc{x} + C$

4. ## Re: indefinite integral of 4cos+sinx-cosx/ sin^2x

Hello, dreamd7358!

$\displaystyle \int \frac{4\cos x+\sin x-\cos x}{\sin^\!}\,dx$

We have: .$\displaystyle \int\frac{3\cos x + \sin x}{\sin^2\!x}\,dx \;=\;\int\left(\frac{\3\cos x}{\sin^2\!x} + \frac{\sin x}{\sin^2\!x}\right)dx$

. . . . . . $\displaystyle =\;\int\left(3\!\cdot\!\frac{1}{\sin x}\!\cdot\!\frac{\cos x}{\sin x} + \frac{1}{\sin x}\right)dx \;=\;\int\left(3\csc x\cot x + \csc x)\,dx$

. . . . . . $\displaystyle =\;-3\csc x + \ln\big|\csc x - \cot x\big|+ C$

5. ## Re: indefinite integral of 4cos+sinx-cosx/ sin^2x

Notice that skeeter did a completely different problem from sakonpure6 and Soroban! dreamd7358, please use parentheses to make you meaning clear. skeeter assumed you meant what you actually wrote: 4cos+sinx-cosx/ sin^2x= 4cos(x)+ sin(x)- (cos(x)/sin^2(x)) while sakonpure6 and Soroban guess that you really meant (4 cos(x)+ sin(x)- cos(x))/sin^2(x).

(Althoug we all had to guess that your initial "4cos+ sinx" meant "4cos(x)+ sin(x)"!