Question

Evaluate $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{dx}}{{1 + \cos x}}}$
A.$- 2$
B.$2$
C.$4$
D.$- 1$

Answer 1

Hint : The integrand is a well-known trigonometric expression. It can be reduced to function which can easily be integrated by using standard formulas of integration. It will also require a method of substitution to evaluate the integral.

Complete step-by-step answer :
The integral which needs to be evaluated is given as
$\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{dx}}{{1 + \cos x}}}$
We know that
$1 + \cos x = 2{\cos ^2}\left( {\dfrac{x}{2}} \right)$
Hence, we have
$\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{dx}}{{1 + \cos x}}} = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{dx}}{{2{{\cos }^2}\left( {\dfrac{x}{2}} \right)}}}$
$\Rightarrow \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{dx}}{{1 + \cos x}}} = _{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}}\smallint \dfrac{{{{\sec }^2}\left( {\dfrac{x}{2}} \right)dx}}{2}$ … (i)
Let $I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{1}{2}{{\sec }^2}\left( {\dfrac{x}{2}} \right)} dx$
Let us put $\dfrac{x}{2} = m$
Differentiating both sides with respect to x, we get
$\dfrac{d}{{dx}}\left( {\dfrac{x}{2}} \right) = \dfrac{{dm}}{{dx}}$

$$\Rightarrow \dfrac{1}{2}\dfrac{{dx}}{{dx}} = \dfrac{{dm}}{{dx}} \\\ \Rightarrow \dfrac{{dx}}{2} = dm \\\$$

Therefore, $I$ becomes
$I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {{{\sec }^2}mdm}$

$$\Rightarrow I = \left[ {{{\tan }^m}} \right]_{\dfrac{\pi }{8}}^{\dfrac{{3\pi }}{8}} \\\ \Rightarrow I = \tan \left( {\dfrac{{3\pi }}{8}} \right) - \tan \left( {\dfrac{\pi }{8}} \right) \\\$$

Let us evaluate the values of $\tan \left( {\dfrac{{3\pi }}{8}} \right)$ and $\tan \left( {\dfrac{\pi }{8}} \right)$ $\tan \left( {\dfrac{\pi }{8}} \right)$

$$\because \tan \left( {\dfrac{{3\pi }}{4}} \right) = \tan \left( {2.\dfrac{{3\pi }}{8}} \right) \\\ \Rightarrow \tan \left( {\dfrac{{3\pi }}{4}} \right) = \dfrac{{2\tan \left( {\dfrac{{3\pi }}{8}} \right)}}{{1 - {{\tan }^2}\left( {\dfrac{{3\pi }}{8}} \right)}} \\\ \Rightarrow - 1 = \dfrac{{2\tan \left( {\dfrac{{3\pi }}{8}} \right)}}{{1 - {{\tan }^2}\left( {\dfrac{{3\pi }}{8}} \right)}} \\\ \Rightarrow - 1 + {\tan ^2}\left( {\dfrac{{3\pi }}{8}} \right) = 2\tan \left( {\dfrac{{3\pi }}{8}} \right) \\\$$ $$\Rightarrow {\tan ^2}\left( {\dfrac{{3\pi }}{8}} \right) - 2\tan \left( {\dfrac{{3\pi }}{8}} \right) - 1 = 0 \\\ \\\$$

Above question is a quadratic equation in $\tan \left( {\dfrac{{3\pi }}{8}} \right)$

$$\therefore \tan \left( {\dfrac{{3\pi }}{8}} \right) = \dfrac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4 \times 1 \times \left( { - 1} \right)} }}{2} \\\ \Rightarrow \tan \left( {\dfrac{{3\pi }}{8}} \right) = \dfrac{{2 \pm \sqrt {4 + 4} }}{2} \\\ \Rightarrow \tan \left( {\dfrac{{3\pi }}{8}} \right) = \dfrac{{2 \pm \sqrt 2 }}{2} \\\ \Rightarrow \tan \left( {\dfrac{{3\pi }}{8}} \right) = 1 \pm \sqrt 2 \\\$$

$\because \dfrac{{3\pi }}{8}$ lies in the first quadrant. Hence $\tan \left( {\dfrac{{3\pi }}{8}} \right)$ has to be positive.
$\therefore \tan \left( {\dfrac{{3\pi }}{8}} \right) = 1 \pm \sqrt 2$
Again,

$$\therefore \tan \left( {\dfrac{\pi }{4}} \right) = \tan \left( {2.\dfrac{\pi }{8}} \right) \\\ \Rightarrow 1 = \dfrac{{2\tan \left( {\dfrac{\pi }{8}} \right)}}{{1 - {{\tan }^2}\left( {\dfrac{\pi }{8}} \right)}} \\\ \Rightarrow 1 - {\tan ^2}\left( {\dfrac{\pi }{8}} \right) = 2\tan \left( {\dfrac{\pi }{8}} \right) \\\ \Rightarrow {\tan ^2}\left( {\dfrac{\pi }{8}} \right) + 2\tan \left( {\dfrac{\pi }{8}} \right) - 1 = 0 \\\$$

Again, the above equation is a quadratic equation in $\tan \left( {\dfrac{\pi }{8}} \right)$

$$\therefore \tan \left( {\dfrac{\pi }{8}} \right) = \dfrac{{ - 2 \pm \sqrt {{{\left( 2 \right)}^2} - 4 \times 1 \times \left( { - 1} \right)} }}{2} \\\ \Rightarrow \tan \left( {\dfrac{\pi }{8}} \right) = \dfrac{{2 \pm \sqrt 8 }}{2} \\\ \Rightarrow \tan \left( {\dfrac{\pi }{8}} \right) = - 1 \pm \sqrt 2 \\\$$

$\because \dfrac{\pi }{8}$ lies in the first quadrant. Hence $\tan \left( {\dfrac{\pi }{8}} \right)$ has to be positive.
$\therefore \tan \left( {\dfrac{\pi }{8}} \right) = - 1 + \sqrt 2$
Hence,

$$I = \tan \left( {\dfrac{{3\pi }}{8}} \right) - \tan \left( {\dfrac{\pi }{8}} \right) \\\ \Rightarrow I = \left( {1 + \sqrt 2 } \right) - \left( { - 1 + \sqrt 2 } \right) \\\ \Rightarrow I = 1 + \sqrt 2 + 1 - \sqrt 2 \\\ I = 2 \\\$$

Thus,
$\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{dx}}{{1 + \cos x}}} = 2$
So, the correct answer is “Option B”.

Note : The students must remember all the basic formulas of trigonometric identities. They must practice to evaluate values of trigonometric functions using simple tricks. One must keep in mind the information of the quadrant in which the angles lie. This helps to identify the correct value of the function from the values obtained after calculation.