1. ## solve equation

Hi everyone,

I would like to know if I did any mistake in my calculation?

Here is the question and my answer:

(a Using equation:
(f(x))n f’(x) dx = 1/n+1(f(x))n+1 + c (n≠-1)

show that
cos x dx = - 1 + c,
(2 + sin x)2 2+ sin x

Where c is an arbitrary constant.

cos x dx = - sin ∫ (x) dx
(2 + sin x)2 (2+sinx)2

= x = - 1
(2+sin x)2 2+sin x + c

2. Originally Posted by valerie-petit
Hi everyone,

I would like to know if I did any mistake in my calculation?

Here is the question and my answer:

(a Using equation:
(f(x))n f’(x) dx = 1/n+1(f(x))n+1 + c (n≠-1)

show that
cos x dx = - 1 + c,
(2 + sin x)2 2+ sin x

Where c is an arbitrary constant.

cos x dx = - sin ∫ (x) dx
(2 + sin x)2 (2+sinx)2

= x = - 1
(2+sin x)2 2+sin x + c

'Tis very hard to read what you have written. So I will work it out, check whether thats what you meant:

$\int \frac{\cos x}{(2 + \sin x)^2} \, dx$

If we choose $f(x) = 2 + \sin x$, then $f'(x) = \cos x$.

Now

$\int \frac{\cos x}{(2 + \sin x)^2} \, dx = \int \frac{f'(x)}{f(x)^2}\, dx = \int f(x)^{-2} \, f'(x)\, dx = \frac{f(x)^{-1}}{-1}+C = -\frac{1}{2+ \sin x} + C$

3. ## Thanks again

Sorry about what I wrote, everytime I do copy/paste from word it does not recognize half of my text.