# Thread: Integral Problem

1. ## Integral Problem

Hi experts,
How does one solve

$-\int\frac{sinx}{cos^2x}dx$

I have attempted to use the identity cos^2x=1/2(1+cos2x) to give
$-\int\frac{sinx}{\frac{1}{2}(1+cos2x)}$
However I am not sure how to proceed any further because of the double angle....

In another attempt, i tried tackling it using the method of substitution such as below
let u=cosx then du = -sinxdx
then
$\int\frac{du}{u^2} = -\frac{1}{u} + c$

But this answer doesnt look right either....

Any tips or suggestions on what I am doing wrong will be appreciated. Thanks
Bugatti

2. Your second solutions looks fine to me, try differentiating the right side and see if you end up with your original function.

3. Originally Posted by bugatti79
In another attempt, i tried tackling it using the method of substitution such as below
let u=cosx then du = -sinxdx
then
$\int\frac{du}{u^2} = -\frac{1}{u} + c$

But this answer doesnt look right either....
The latter is correct

4. ## Integral Problem

Thanks guys,

I should have been more thorough and checked by differentiating back.
cheers
bugatti79

5. Once You realize that is...

$\frac{d}{dx} \cos x = - \sin x$ (1)

... the indefinite integral...

$\displaystyle - \int \frac{\sin x}{\cos^{2} x}\ dx$ (2)

... becomes incredibly easy ...

Kind regards

$\chi$ $\sigma$