# Is this integral evaluation valid?

#### Bmanmcfly

##### Member
[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.

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#### MarkFL

Staff member
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

##### Member
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.