Question

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

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

Answer 1

The correct option is A $\frac{1}{2}$

$\int \frac{1}{1-{\cos }^{4}x}dx=-\frac{1}{2\tan x}+\frac{k}{\sqrt{2}}{\tan }^{-1}\left(\frac{\tan x}{\sqrt{2}}\right)+C$
Consider, $\int \frac{1}{1-{\cos }^{4}x}dx$
$=\int \frac{1}{{\sin }^{2}x(1+{\cos }^{2}x)}dx$
$=\int \frac{{\sec }^{4}x}{{\tan }^{2}x({\sec }^{2}x+1)}dx$
$=\int \frac{(1+{\tan }^{2}x){\sec }^{2}x}{{\tan }^{2}x({\tan }^{2}x+2)}dx$
Put $\tan x=t$
${\sec }^{2}xdx=dt$
$I=\int \frac{(1+t^2)}{t^2(t^2+2)}dt$ ....(1)
Resolving $\frac{(1+t^2)}{t^2(t^2+2)}$ into partial fractions
$\frac{1+t^2}{t^2(t^2+2)}=\frac{A}{t}+\frac{B}{t^2}+\frac{Ct+D}{t^2+2}$ ....(2)
$\Rightarrow 1+t^2=At(t^2+2)+B(t^2+2)+(Ct+D)t^2$
$\Rightarrow B=\frac{1}{2};A=0;C=0;D=\frac{1}{2}$
Substituting these values in (2), we get
$\frac{1+t^2}{t^2(t^2+2)}=\frac{1}{{2t}^2}+\frac{1}{{2(t}^2+2)}$
Using this in (1), we get
$\int \frac{(1+t^2)}{t^2(t^2+2)}dt=\frac{1}{2}\int \frac{1}{t^2}dt+\frac{1}{2}\int \frac{1}{{(t}^2+2)}dt$
$=-\frac{1}{2}\frac{1}{t}+\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{t}{\sqrt{2}}+C$
$I=-\frac{1}{2}\frac{1}{\tan x}+\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\tan x}{\sqrt{2}}\right)+C$
Comparing with given , we get
$k=\frac{1}{2}$