Question

The value of $\int_0^{\dfrac{\pi }{2}} {{{\cos }^2}xdx} $ is
A. 0
B. $\dfrac{\pi }{6}$
C. $\dfrac{\pi }{4}$
D. $\dfrac{\pi }{3}$

Answer 1

Hint: Start by converting the ${\cos ^2}x$ in the form of linear angles like $\cos ax$ as we don’t have the direct formula for integration. After converting the given integral apply the integral values to calculate the value of it.

Complete step-by-step answer:

Let $l = \int_0^{\dfrac{\pi }{2}} {{{\cos }^2}xdx} $
We can expand and write ${\cos ^2}x = \dfrac{{\cos 2x + 1}}{2}$, therefore,
$l = \int_0^{\dfrac{\pi }{2}} {\left[ {\dfrac{{\cos 2x + 1}}{2}} \right]dx} $
Next step is to take out the constants common,
$l = \dfrac{1}{2}\left[ {\int_0^{\dfrac{\pi }{2}} {\cos 2xdx + \int_0^{\dfrac{\pi }{2}} 1 dx} } \right]$
Now, let us perform integration,
$l = \dfrac{1}{2}\left[ {\dfrac{{\sin 2x}}{2}} \right]_0^{\dfrac{\pi }{2}} + \dfrac{1}{2}\left[ x \right]_0^{\dfrac{\pi }{2}}\left[ {\because \int {\cos axdx = \dfrac{{\sin ax}}{a},\int {1dx = x} } } \right]$
On applying the limits, we get,
$l = \dfrac{1}{4}\left[ {\sin \pi - \sin 0} \right] + \dfrac{1}{2}\left[ {\dfrac{\pi }{2} - 0} \right]\left( {\because \sin \pi = \sin 0 = 0} \right)$
On putting the values, we get,
$l = \dfrac{1}{4}\left[ {0 - 0} \right] + \dfrac{\pi }{4}$
On simplifying, we get,
$l = \dfrac{\pi }{4}$
Option C is the answer.

Note: In these types of questions, we need to convert the integral into simple expansions such that we can apply direct integration on them just by putting the formulae. So the basic conversions and formulae of both trigonometry and integrals need to be known for solving the given question.