Question
Find the integrals of the functions in Exercises:
$\frac{\sin^2\text{x}}{1+\cos\text{x}}$
$\frac{\sin^2\text{x}}{1+\cos\text{x}}$
✓
Answer
$\frac{\sin^2\text{x}}{1+\cos\text{x}}=\frac{\bigg(2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}\bigg)^2}{2\cos^2\frac{\text{x}}{2}}$ $\ \ \ \ \ \bigg[\sin\text{x}=2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2};\cos\text{x}=2\cos^2\frac{\text{x}}{2}-1\bigg]$ $=\frac{4\sin^2\frac{\text{x}}{2}\cos^2\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}$ $=2\sin^2\frac{\text{x}}{2}$ $=1-\cos\text{x}$ $\therefore\int\frac{\sin^2\text{x}}{1+\cos\text{x}}\text{ dx}=\int(1-\cos\text{x})\text{ dx}$ $=\text{x}-\sin\text{x}+\text{C}$
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