MHB Is this integral evaluation valid?

[bmanmcfly]

[SOLVED]Is this integral evaluation valid?

Hi, so I started with [math]\int \frac{\sin(x)\cos^2(x)}{5+\cos^2(x)}dx[/math]I made u=cos(x) dx=sin(x) leaving [math]\int \frac{u^2}{5+u^2}dx[/math]At this point I was thinking that it looked like an inverse tan, but I was lazy, so instead I tried [math]\int\frac{u^2}{5}dx+\int\frac{u^2dx}{u^2}[/math]In the name of brevity, I concluded with [math]\frac{\cos^3(x)}{15}+ \cos(x) + C[/math]was this a valid way to perform the integration, or should I have went with partial fractions instead? Or just stuck with the inverse tan?Thanks.

 

[MarkFL]

I would be careful with the notation and differentiation. Using the substitution:

$$u=\cos(x)\,\therefore\,du=-\sin(x)\,dx$$

and the integral becomes:

$$-\int\frac{u^2}{u^2+5}\,du$$

Next, you have tried to use:

$$\frac{a}{a+b}=\frac{a}{a}+\frac{a}{b}$$

and this simply is not true. I would suggest rewriting the integrand as:

$$\frac{(u^2+5)-5}{u^2+5}=1-\frac{5}{u^2+5}$$

Now you may integrate term by term, then back-substitute for $u$.

 

[bmanmcfly]

I would be careful with the notation and differentiation. Using the substitution:

$$u=\cos(x)\,\therefore\,du=-\sin(x)\,dx$$

and the integral becomes:

$$-\int\frac{u^2}{u^2+5}\,du$$

Next, you have tried to use:

$$\frac{a}{a+b}=\frac{a}{a}+\frac{a}{b}$$

and this simply is not true. I would suggest rewriting the integrand as:

$$\frac{(u^2+5)-5}{u^2+5}=1-\frac{5}{u^2+5}$$

Now you may integrate term by term, then back-substitute for $u$.

Oops, forgot te minus here, not written down...

Ok, thought I was doing that too simply.

Thanks.