całki

[margosia22]

\(\int\frac{sinx}{3-2cosx}dx\)
\(\int\frac{1}{7+sinx}dx\)

Proszę o pomoc

[irena]

a)
\(\int\frac{sinx}{3-2cosx}dx\\\(3-2cosx=t\\2sinxdx=dt\\sinxdx=\frac{1}{2}dt\)\\\int(\frac{1}{2t})dt=\frac{1}{2}\int\frac{1}{t}dt=\frac{1}{2}ln|t|=\frac{1}{2}ln(3-2cosx)+C\)

[irena]

b)
\(\int\frac{1}{7+sinx}dx=\\\(tg\frac{x}{2}=t\\\frac{x}{2}=arc\ tg\ t\\x=2arc\ tg\ t\\dx=\frac{2}{1+t^2}dt\\sinx=\frac{2t}{1+t^2}\)\\=\int\frac{1}{7+\frac{2t}{1+t^2}}\cdot\frac{2}{1+t^2}dt=\int\frac{2}{7+7t^2+2t}dt=\frac{2}{7}\int\frac{1}{t^2+\frac{2}{7}t+1}dt=\frac{2}{7}\int\frac{1}{(t+\frac{1}{7})^2+\frac{48}{49}}dt=\\\(t+\frac{1}{7}=q\\dt=dq\)\\=\frac{2}{7}\int\frac{1}{q^2+\frac{48}{49}}dq=\\\(\frac{q}{\sqrt{\frac{48}{49}}}=p\\q=\sqrt{\frac{48}{49}}p\\dq=\sqrt{\frac{48}{49}}dp\)\\=\frac{2}{7}\int\frac{1}{\frac{48}{49}p^2+\frac{48}{49}}\cdot\sqrt{{\frac{48}{49}}}dp=\frac{2}{7}\cdot\sqrt{\frac{49}{48}}\int\frac{1}{p^2+1}dp=\frac{2}{7}\cdot\frac{7}{4\sqrt{3}}arc\ tg\ p=\\=\frac{1}{2\sqrt{3}}arc\ tg\ \frac{7}{4\sqrt{3}}q=\\=\frac{1}{2\sqrt{3}}arc\ tg\ \frac{7}{4\sqrt{3}}(t+\frac{1}{7})=\\=\frac{\sqrt{3}}{6}arc\ tg\ \frac{7t+1}{4\sqrt{3}}=\\=\frac{\sqrt{3}}{6}\ arc\ tg\ \frac{7tg\frac{x}{2}+1}{4\sqrt{3}}+C\)

P. S. Może jest jakiś prostszy sposób- mniej podstawień, ale sprawdzałam wynik i jest chyba dobrze