Question

Evaluate: $\underset{0}{\overset{\pi /2}{\int }}\frac{dx}{1+2\mathrm{sinx}+cosx}$qual

• A. $\left(1/2\right)In3$
• B. $In3$
• C. $\left(4/3\right)In3$
• D. none of these

Answer 1

Answer: (A)

$\left(1/2\right)In3$

$I={\int }_0^{\frac{\pi }{2}}\frac{1}{1+2sinx+cosx}dx$

$I={\int }_0^{\frac{\pi }{2}}\frac{1}{1+\frac{4tan\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}+\frac{1-ta{n}^2\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}}dx$

$\left[\begin{array}{c}\because sin2x=\frac{2tanx}{1+ta{n}^{2}x}\end{array}\right]$ & $\left[\begin{array}{c}cos2x=\frac{1-ta{n}^{2}x}{1+ta{n}^{2}x}\end{array}\right]$

$I={\int }_0^{\frac{\pi }{2}}\frac{1+ta{n}^2\frac{x}{2}}{1+ta{n}^2\frac{x}{2}+4tan\frac{x}{2}+1-ta{n}^2\frac{x}{2}}dx$

$I={\int }_0^{\frac{\pi }{2}}\frac{se{c}^{2}x}{2+4tan\frac{x}{2}}dx$ $\to$ $I={\int }_0^{\frac{\pi }{2}}\frac{se{c}^{2}x}{2(1+2tan\frac{x}{2})}dx$

put $1+2tan\frac{x}{2}=t\to 2\times \frac{1}{2}se{c}^{2}xdx=dt$

also

when $x=\frac{\pi }{2}\to t=3andx=0\to t=1$

$I={\int }_1^3\frac{dt}{2t}dx$

$\Rightarrow \frac{1}{2}{\left[\begin{array}{c}\ln t\end{array}\right]}_1^3$

$\Rightarrow \left(\frac{1}{2}\right)\ln 3-\frac{1}{2}\ln 1$

$=\frac{1}{2}\ln 3$

hence correct answer is option $A$