$\displaystyle\int\dfrac{\sin^3 \frac{x}{2}}{\cos \frac{x}{2} \sqrt{\cos x+\cos^2 x+\cos^3 x}}dx=\dfrac{1}{2}.\displaystyle\int\dfrac{2.sin^2 \frac{x}{2}\;2.sin\frac{x}{2}.cos\frac{x}{2}}{2cos^2\frac{x}{2} \sqrt{cos x+cos^2 x+cos^3 x}}dx$
$\boxed{\text{Kural:}\quad cosx=2cos^2\frac{x}{2}-1\;=\;1-2.sin^2\frac{x}{2}}$
$\dfrac{1}{2}.\displaystyle\int\dfrac{2.sin^2 \frac{x}{2}\;2.sin\frac{x}{2}.cos\frac{x}{2}}{2cos^2\frac{x}{2} \sqrt{cos x+cos^2 x+cos^3 x}}dx=\dfrac{1}{2}.\displaystyle\int\dfrac{(1-cosx)sinx.dx}{(1+cosx)\sqrt{cos x+cos^2 x+cos^3 x}}$
$cosx=j$ dönüşümü yaparsak
$sinx.dx=-dj$ olur ve ifadeyi düzenlersek
$\dfrac{1}{2}.\displaystyle\int\dfrac{(1-cosx)sinx.dx}{(1+cosx)\sqrt{cos x+cos^2 x+cos^3 x}}=\dfrac{1}{2}.\displaystyle\int\dfrac{(1-j)(-dj)}{(1+j)\sqrt{j+j^2+j^3}}$ her tarafı $i+j$ ile çarpalım,
$\dfrac{1}{2}.\displaystyle\int\dfrac{(1-j)(-dj)}{(1+j)\sqrt{j+j^2+j^3}}=-\dfrac{1}{2}.\displaystyle\int\dfrac{1-j^2}{(1+2j+j^2)\sqrt{j+j^2+j^3}}dj$
her tarafı $j^2$ ye bölelim ,amacımız bir forma benzetmek.
bu "forma" değil "form" olan forma .
$-\dfrac{1}{2}.\displaystyle\int\dfrac{1-j^2}{(1+2j+j^2)\sqrt{j+j^2+j^3}}dj=\dfrac{1}{2}.\displaystyle\int\dfrac{\left(1-\frac{1}{j^2}\right)}{(\frac{1}{j}+2+j)\sqrt{\frac{1}{j}+1+j}}dj$
şimdi ise,
$\boxed{\left(j+\frac{1}{j}+j\right)=u^2}$ dönüşümü uygulayalım,
$\boxed{\left(1-\dfrac{1}{j^2}\right)dj=2u.du}$ olur, ne kadar ahenkli değil mi?
$\dfrac{1}{2}.\displaystyle\int\dfrac{2u.du}{(u^2+1)\sqrt{u^2}}=arctanu+C$
$\boxed{\boxed{arctanu+C=arctan\left(\sqrt{j+\frac{1}{j}+j}\right)+C=\displaystyle \displaystyle I = \tan^{-1}\sqrt{\left(\cos x+\sec x+1\right)}+C}}$