Question

$\int \frac{\sin 2x}{2+3{\cos }^{2}x}dx=$

• A. $-\frac{1}{3}\log |2+3{\cos }^{2}x|+c$
• B. $\log |2+3{\cos }^{2}x|+c$
• C. $\frac{1}{3}\log |2+3{\cos }^{2}x|+c$
• D. $-\log |2+3{\cos }^{2}x|+c$

Answer 1

The correct option is A $-\frac{1}{3}\log |2+3{\cos }^{2}x|+c$

$\int \frac{sin2x}{2+3co{s}^{2}x}dx$

Substitute $co{s}^{2}x=t$

$\Rightarrow 2cosx(-sinx)dx=dt$

$\Rightarrow sin2xdx=-dt$

$\int \frac{-dt}{2+3t}=-\frac{1}{3}log(2+3t)+c$

$=-\frac{1}{3}log(2+3co{s}^{2}x)+c$

$\int \frac{\sin 2x}{2+3{\cos }^{2}x}dx=-\frac{1}{3}\log (2+3{\cos }^{2}x)+c$