Question

$\int \frac{\sin 2x}{1+{\cos }^{2}x}dx$

• A. $\ln (1-{\cos }^{2}x)+C$
• B. $-\ln (1+{\sin }^{2}x)+C$
• C. $-\ln (1+{\cos }^{2}x)+C$
• D. $\ln \left(\frac{1}{1-{\cos }^{2}x}\right)+C$

Answer 1

The correct option is C $-\ln (1+{\cos }^{2}x)+C$

$\int \frac{\sin 2x}{1+{\cos }^{2}x}dx$

Put $1+{\cos }^{2}x=t0+2\cos (-\sin x)dx=dt-2\sin x\cos xdx=dt-\sin 2xdx=dt\sin 2xdx=-dt$

Thus $\int \frac{\sin 2x}{1+{\cos }^{2}x}dx=\int \frac{-dt}{t}=-\int \frac{dt}{t}=-\log \left|t\right|+c=-\log (1+{\cos }^{2}x)+c$