Integrate root cos2x divided by sinx with respect to x

\displaystyle \int{\frac{\sqrt{\cos 2x}}{\sin x}dx}

\displaystyle =\int{\frac{\sqrt{\frac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}}}{\sin x}}\,\,dx

\displaystyle =\int{\frac{\sqrt{1-{{\tan }^{2}}x}}{\sec x.\sin x}\,dx}

\displaystyle =\int{\frac{\sqrt{1-{{\tan }^{2}}x}}{\tan x}\,\,dx}

\displaystyle =\int{\frac{\sqrt{1-{{\tan }^{2}}x}}{\tan x\left( 1+{{\tan }^{2}}x \right)}\,{{\sec }^{2}}x\,\,dx}

\displaystyle =\int{\frac{\sqrt{1-{{\tan }^{2}}x}}{{{\tan }^{2}}x\left( 1+{{\tan }^{2}}x \right)}}\,\,\tan x{{\sec }^{2}}x\,\,dx

Let \displaystyle \sqrt{1-{{\tan }^{2}}x}\,\,=t

\displaystyle 1-{{\tan }^{2}}x={{t}^{2}}

Differentiate both sides w.r.t x

\displaystyle -2\tan x{{\sec }^{2}}x\,\,dx\,\,=2\,tdt

\displaystyle \Rightarrow I=-\int{\frac{t}{\left( 1-{{t}^{2}} \right)\left( 2-{{t}^{2}} \right)}\,t\,\,dt}

\displaystyle =-\int{\frac{{{t}^{2}}}{\left( 1-{{t}^{2}} \right)\left( 2-{{t}^{2}} \right)}\,\,dt}

Let \displaystyle \frac{{{t}^{2}}}{\left( 1-{{t}^{2}} \right)\left( 2-{{t}^{2}} \right)}=\frac{At+B}{1-{{t}^{2}}}\,\,+\,\frac{Ct+D}{2-{{t}^{2}}}

\displaystyle \begin{array}{l}\Rightarrow \frac{{{t}^{2}}}{\left( 1-{{t}^{2}} \right)\left( 2-{{t}^{2}} \right)}=\frac{At\left( 2-{{t}^{2}} \right)+B\left( 2-{{t}^{2}} \right)+Ct{{\left( 1-t \right)}^{2}}+D\left( 1-{{t}^{2}} \right)}{\left( 1-{{t}^{2}} \right)\left( 2-{{t}^{2}} \right)}\\\Rightarrow {{t}^{2}}=2At-A{{t}^{2}}+2B-B{{t}^{2}}+Ct-C{{t}^{3}}+D-D{{t}^{2}}\\\Rightarrow {{t}^{2}}=-{{t}^{3}}\left( A+C \right)-{{t}^{2}}\left( B+D \right)+\left( 2A+C \right)t+\left( 2B+D \right)\end{array}

Equating coefficient both sides, we get

\displaystyle \begin{array}{l}A+C=0\\B+D=-1\\2A+C=0\\2B+D=0\end{array}

On solving above equations

\displaystyle \begin{array}{l}A=0\\C=0\\B=1\\D=-2\end{array}

\displaystyle \therefore I=\int{\frac{1}{1-{{t}^{2}}}}dt-2\,\,\int{\frac{1}{2-{{t}^{2}}}\,dt}

\displaystyle \begin{array}{l}=\frac{1}{2}\,\,\log \left| \frac{1+t}{1-t} \right|-\frac{2}{2\sqrt{2}}\,\,\log \left| \frac{\sqrt{2}+t}{\sqrt{2}-t} \right|+C\\=\frac{1}{2}\log \left| \frac{1+\sqrt{1-{{\tan }^{2}}}x}{1-\sqrt{1-{{\tan }^{2}}x}} \right|-\frac{1}{\sqrt{2}}\,\log \left| \frac{\sqrt{2}+\sqrt{1-{{\tan }^{2}}}x}{\sqrt{2}-\sqrt{1-{{\tan }^{2}}}x} \right|+C\end{array}

Post Author: E-Maths

  • nidhi sharma

    You forgot negative sign (which was there in the integral I) after partial fractions.