Question

$\int \frac{dx}{4{\sin }^{2}x+5{\cos }^{2}x}=k{\tan }^{-1}\left(\frac{2}{\sqrt{5}}\tan x\right)+c$, then $k=$

• A. $\frac{1}{\sqrt{5}}$
• B. $\frac{\sqrt{5}}{2}$
• C. $\frac{1}{2\sqrt{5}}$
• D. $\frac{2}{\sqrt{5}}$

Answer 1

Answer: (C)

$\frac{1}{2\sqrt{5}}$

$\int \frac{dx}{4si{n}^{2}x+5co{s}^{2}x}$ dividing N(1) and D(x) by $co{s}^{2}x.$
$\int \frac{se{c}^{2}xdx}{4ta{n}^{2}x+5}lettanx=t$
$\frac{1}{2}\int \frac{d\left(2t\right)}{(2t{)}^2+(\sqrt{5}{)}^2}=\frac{1}{2}\left({}^1{/}_{\sqrt{5}}\right){\tan }^{-1}\left(\frac{2t}{\sqrt{5}}\right)+c$
$=\frac{1}{2\sqrt{5}}{\tan }^{-1}\left(\frac{2tanx}{\sqrt{5}}\right)+c$
$k=\frac{1}{2\sqrt{5}}$