Is this integral evaluation valid?

[bmanmcfly]

[SOLVED]Is this integral evaluation valid?

Hi, so I started with \(\displaystyle \int \frac{\sin(x)\cos^2(x)}{5+\cos^2(x)}dx\)I made u=cos(x) dx=sin(x) leaving \(\displaystyle \int \frac{u^2}{5+u^2}dx\)At this point I was thinking that it looked like an inverse tan, but I was lazy, so instead I tried \(\displaystyle \int\frac{u^2}{5}dx+\int\frac{u^2dx}{u^2}\)In the name of brevity, I concluded with \(\displaystyle \frac{\cos^3(x)}{15}+ \cos(x) + C\)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:

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

and the integral becomes:

\(\displaystyle -\int\frac{u^2}{u^2+5}\,du\)

Next, you have tried to use:

\(\displaystyle \frac{a}{a+b}=\frac{a}{a}+\frac{a}{b}\)

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

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

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:

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

and the integral becomes:

\(\displaystyle -\int\frac{u^2}{u^2+5}\,du\)

Next, you have tried to use:

\(\displaystyle \frac{a}{a+b}=\frac{a}{a}+\frac{a}{b}\)

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

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

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.