Question

TT/6
Cos[x]
dx
(1 + Sin[x]) (2 - Sin[x])
0
O (log(3))/2
O log3)/3
O (log2)/5
(log2)/3​

Answer 1

Given,

$\[\begin{array}{l}\int_0^{\frac{\pi }{6}} {\frac{{\cos x}}{{\left( {1 + \sin x} \right)\left( {2 - \sin x} \right)}}} dx\\\left( a \right)\frac{{\log 3}}{2}\\\left( b \right)\frac{{\log 3}}{3}\\\left( c \right)\frac{{\log 2}}{5}\\\left( d \right)\frac{{\log 2}}{3}\end{array}\]$

Solution,

$\[ = \int_0^{\frac{\pi }{6}} {\frac{{\cos x}}{{\left( {1 + \sin x} \right)\left( {2 - \sin x} \right)}}} dx\]$----------(1)

Assume,

$\[\begin{array}{l}1 + \sin x = t\\\cos xdx = dt\end{array}\]$----------(2)

Put this value in equation 1,

$\[ = \int_1^{\frac{3}{2}} {\frac{{dt}}{{t\left( {3 - t} \right)}}} \]$

$\[ = \int_1^{\frac{3}{2}} {\frac{1}{3}\left( {\frac{1}{t} - \frac{1}{{3 - t}}} \right)dt} \]$

$\[ = \mathop {\frac{1}{3}\left( {\ln t - \ln \left( {3 - t} \right)} \right)}\nolimits_1^{\frac{3}{2}} \]$

$\[ = \mathop {\frac{1}{3}\ln \left( {\frac{t}{{3 - t}}} \right)}\nolimits_1^{\frac{3}{2}} \]$

Put the limit values,

$\[ = \frac{1}{3}\left\{ {\ln \left( {\frac{{\frac{3}{2}}}{{3 - \frac{3}{2}}}} \right) - \ln \left( {\frac{1}{{3 - 1}}} \right)} \right\}\]$

$\[ = \frac{1}{3}\left\{ {\ln 1 - \ln \left( {\frac{1}{2}} \right)} \right\}\]$

$\[ = \frac{1}{3}\left\{ {0 + \ln 2} \right\}\]$

$\[ = \frac{{\ln 2}}{3}\]$

Hence the correct option is (d), i.e. $\[\frac{{\ln 2}}{3}\]$.