# Evaluate the following definite integral :

\[\int\limits_0^{\dfrac{\pi }{2}} {{{\cos }^2}xdx} \]

Answer

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Hint :- Use the formula \[\;(cos2x = 2co{s^2}x - 1)\]

Let \[I = \int\limits_0^{\dfrac{\pi }{2}} {{{\cos }^2}xdx} \]

$ = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{1 + \cos 2x}}{2}dx} $ ………………..using \[\;(cos2x = 2co{s^2}x - 1)\]

$ = \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{2}} {1 + \cos 2x dx} $

$ = \dfrac{1}{2}\left( {\int\limits_0^{\dfrac{\pi }{2}} {1dx + \int\limits_0^{\dfrac{\pi }{2}} {\cos 2xdx} } } \right)$

$ = \dfrac{1}{2}\left[ {x + \dfrac{{\sin 2x}}{2}} \right]_0^{\dfrac{\pi }{2}}$………….as we know \[\left( {\int {\cos 2xdx = \dfrac{{\sin 2x}}{2}} } \right)\]

$ = \dfrac{1}{2}\left[ {\left( {\dfrac{\pi }{2} + \dfrac{{\sin \pi }}{2}} \right) - \left( {0 + \dfrac{{\sin 0}}{2}} \right)} \right] = \dfrac{\pi }{4}$

Note :- Try to make the integral in simple form as you can apply general formula to get the result

of integration.

Let \[I = \int\limits_0^{\dfrac{\pi }{2}} {{{\cos }^2}xdx} \]

$ = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{1 + \cos 2x}}{2}dx} $ ………………..using \[\;(cos2x = 2co{s^2}x - 1)\]

$ = \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{2}} {1 + \cos 2x dx} $

$ = \dfrac{1}{2}\left( {\int\limits_0^{\dfrac{\pi }{2}} {1dx + \int\limits_0^{\dfrac{\pi }{2}} {\cos 2xdx} } } \right)$

$ = \dfrac{1}{2}\left[ {x + \dfrac{{\sin 2x}}{2}} \right]_0^{\dfrac{\pi }{2}}$………….as we know \[\left( {\int {\cos 2xdx = \dfrac{{\sin 2x}}{2}} } \right)\]

$ = \dfrac{1}{2}\left[ {\left( {\dfrac{\pi }{2} + \dfrac{{\sin \pi }}{2}} \right) - \left( {0 + \dfrac{{\sin 0}}{2}} \right)} \right] = \dfrac{\pi }{4}$

Note :- Try to make the integral in simple form as you can apply general formula to get the result

of integration.

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