# Thread: Primitives to trigonometric expression

1. ## Primitives to trigonometric expression

By using substitution t=tan(x/2), find all primitives to:
$$\frac{1}{(sin(x))^3}$$
My attempt:
$$dx=\frac{2}{t^2+1}$$
$$sin(x)=\frac{2t}{1+t^2}$$
$$\int\frac{1}{(\frac{2t}{1+t^2})^3}*\frac{2}{t^2+ 1}dt$$
$$=\frac{1}{4}\int\frac{t^4+2t^2+1}{t^3}dt$$
After polynomial devision
$$\frac{1}{4}(\int t + \int\frac{2t^2+1}{t^3}) = \frac{t^2}{8}+\frac{\ln(t)}{2}-\frac{1}{8t^2}$$

The right anwer is:
$$-\frac{cosx}{2(sinx)^2}+\frac{1}{2}\ln{(tan(x/2))}+C$$

2. ## Re: Primitives to trigonometric expression

You haven't attempted to convert your $t$s baxk into $x$s.

3. ## Re: Primitives to trigonometric expression

$$=\frac{tan^4(\frac{x}{2})-1}{8tan^2(\frac{x}{2})} + \frac{\ln(tan(\frac{x}{2}))}{2}+C$$ ??

4. ## Re: Primitives to trigonometric expression

$\displaystyle \frac{t^2}{8}-\frac{1}{8t^2}=\frac{t^4-1}{8t^2}=-\frac{(1+t^2)(1-t^2)}{8t^2}$

We know $\displaystyle \cos{x}=\frac{1-t^2}{1+t^2}$ and $\displaystyle \sin{x}=\frac{2t}{1+t^2}$, so try to collect terms that look like that:

$\displaystyle =-\frac{1+t^2}{2t}\frac{(1-t^2)}{4t}=-\frac{1+t^2}{2t}\frac{1-t^2}{1+t^2}\frac{1+t^2}{2t}\frac{1}{2}=-\frac{\cos{x}}{2\sin^2{x}}$

Alternatively, you could plug in the reverse substitutions:

$\displaystyle \tan{\frac{x}{2}}=\frac{\sin{x}}{1+\cos{x}}$

$\displaystyle \tan{\frac{x}{2}}=\frac{1-\cos{x}}{\sin{x}}$

- Hollywood