# Thread: Integrate sinx - 2cosx

1. ## Integrate sinx - 2cosx

I am trying to use substitution on this indefinite integral.

Here is what I have so far:

$\displaystyle \int sinx - 2cosx dx$

let u = -cosx

du/dx = sinx

du = sinx dx - so I am left with:

$\displaystyle \int 2u du$

$\displaystyle = [u^2]$

sub back in

$\displaystyle - cos^2 x$

Where am I going wrong?

Angus

2. ## Re: Integrate sinx - 2cosx

what you want to integrate is

$\displaystyle \int (\sin(x)-2\cos(x))\;dx$

But you are integrating

$\displaystyle \int -2\sin(x)\cos(x)\;dx$

3. ## Re: Integrate sinx - 2cosx

Originally Posted by angypangy
I am trying to use substitution on this indefinite integral.

Here is what I have so far:

$\displaystyle \int sinx - 2cosx dx$
$\displaystyle = \int sin x dx- 2\int cos x dx$

Or did you mean $\displaystyle \int sin(x)(-2cos(x))dx$ as issacnewton suggests?

let u = -cosx

du/dx = sinx

du = sinx dx - so I am left with:

$\displaystyle \int 2u du$

$\displaystyle = [u^2]$

sub back in

$\displaystyle - cos^2 x$

Where am I going wrong?

Angus

4. ## Re: Integrate sinx - 2cosx

Originally Posted by issacnewton
what you want to integrate is

$\displaystyle \int (\sin(x)-2\cos(x))\;dx$

But you are integrating

$\displaystyle \int -2\sin(x)\cos(x)\;dx$

Thanks Isaac

I was revising and read up on substitution assuming I would need to substitute. But of course you are correct. Thanks.