Question

$\int \frac{dx}{2-3\cos 2x}=$

• A. $\frac{1}{\sqrt{5}}\log |\frac{\sqrt{5}tanx-1}{\sqrt{5}tanx+1}|+c$
• B. $\frac{1}{2\sqrt{5}}\log |\frac{\sqrt{5}tanx-1}{\sqrt{5}tanx+1}|+c$
• C. $\frac{-1}{2\sqrt{5}}\log |\frac{\sqrt{5}tanx-1}{\sqrt{5}tanx+1}|+c$
• D. $-\frac{1}{\sqrt{5}}\log |\frac{\sqrt{5}tanx-1}{\sqrt{5}tanx+1}|+c$

Answer 1

Answer: (B)

$\frac{1}{2\sqrt{5}}\log |\frac{\sqrt{5}tanx-1}{\sqrt{5}tanx+1}|+c$

$\int \frac{dx}{2-3(1-2si{n}^{2}x)}=\int \frac{dx}{6si{n}^{2}x-1}$

dividing neumerator denominator by $se{c}^{2}x$

$\int \frac{se{c}^{2}xdx}{5ta{n}^{2}x-se{c}^{2}x}=\int \frac{se{c}^{2}xdx}{5ta{n}^{2}x-1}{=}^1{/}_2\int \frac{dta{n}^x}{5ta{n}^{2}x-1}$

${}^1{/}_{2\sqrt{5}}\int \frac{d\left(\sqrt{5}tanx\right)}{(\sqrt{5}tanx{)}^2-1}={}^1{/}_{2\sqrt{5}}log\left|\frac{\sqrt{5}tanx-1}{\sqrt{5}tanx+1}\right|$

$={}^1{/}_{2\sqrt{5}}log\left|\frac{\sqrt{5}tanx-1}{\sqrt{5}tanx+1}\right|+c$