Question

If \[I = \int\limits_0^{\dfrac{\pi }{4}} {{{\sin }^2}xdx} \] and \[J = \int\limits_0^{\dfrac{\pi }{4}} {{{\cos }^2}xdx} \], then what is the value of \[I\]?
A. \[\dfrac{\pi }{4} - J\]
B. \[2J\]
C. \[J\]
D. \[\dfrac{J}{2}\]

Answer 1

Hint: Here, 2 definite integrals are given. First, add both given integrals and simplify them by applying the trigonometric identity \[{\sin ^2}x + {\cos ^2}x = 1\]. Then, solve the integral by using the integration formula \[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b\]. After that, apply the limits. In the end, simplify the equation to get the required answer.



Formula Used:\[{\sin ^2}x + {\cos ^2}x = 1\]
\[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b\]
Integration rule: \[\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx + \int\limits_a^b {g\left( x \right)} dx\]



Complete step by step solution:The given definite integrals are \[I = \int\limits_0^{\dfrac{\pi }{4}} {{{\sin }^2}xdx} \] and \[J = \int\limits_0^{\dfrac{\pi }{4}} {{{\cos }^2}xdx} \].

Now add both integrals.
\[I + J = \int\limits_0^{\dfrac{\pi }{4}} {{{\sin }^2}xdx} + \int\limits_0^{\dfrac{\pi }{4}} {{{\cos }^2}xdx} \]
Apply the sum rule of the integration \[\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx + \int\limits_a^b {g\left( x \right)} dx\] on the right-hand side.
\[I + J = \int\limits_0^{\dfrac{\pi }{4}} {\left[ {{{\sin }^2}x + {{\cos }^2}x} \right]dx} \]
Apply the trigonometric identity \[{\sin ^2}x + {\cos ^2}x = 1\].
\[I + J = \int\limits_0^{\dfrac{\pi }{4}} {1dx} \]
Solve the integral by using the integration formula \[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b\].
\[I + J = \left[ x \right]_0^{\dfrac{\pi }{4}}\]
Apply the upper and lower limits.
\[I + J = \left[ {\dfrac{\pi }{4} - 0} \right]\]
\[ \Rightarrow I + J = \dfrac{\pi }{4}\]
\[ \Rightarrow I = \dfrac{\pi }{4} - J\]



Option ‘A’ is correct



Note: The other way to solve the required integral \[I = \int\limits_0^{\dfrac{\pi }{4}} {{{\sin }^2}xdx} \] is:
 \[I = \int\limits_0^{\dfrac{\pi }{4}} {{{\sin }^2}xdx} \]
Simplify the term using the trigonometric identity \[{\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}\].
\[I = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{1 - \cos 2x}}{2}dx} \]
\[ \Rightarrow I = \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{4}} {\left( {1 - \cos 2x} \right)dx} \]
Apply the sum rule of integration \[\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx + \int\limits_a^b {g\left( x \right)} dx\].
\[ \Rightarrow I = \dfrac{1}{2}\left[ {\int\limits_0^{\dfrac{\pi }{4}} {1dx} - \int\limits_0^{\dfrac{\pi }{4}} {\cos 2xdx} } \right]\]
\[ \Rightarrow I = \dfrac{1}{2}\left[ {x - \dfrac{1}{2}\sin 2x} \right]_0^{\dfrac{\pi }{4}}\]
Apply the limits.
\[ \Rightarrow I = \dfrac{1}{2}\left[ {\left( {\dfrac{\pi }{4} - \dfrac{1}{2}\sin 2\dfrac{\pi }{4}} \right) - \left( {0 - \dfrac{1}{2}\sin 2\left( 0 \right)} \right)} \right]\]
\[ \Rightarrow I = \dfrac{1}{2}\left[ {\left( {\dfrac{\pi }{4} - \dfrac{1}{2}\sin \dfrac{\pi }{2}} \right) - \left( {0 - 0} \right)} \right]\]
\[ \Rightarrow I = \dfrac{1}{2}\left( {\dfrac{\pi }{4} - \dfrac{1}{2}\left( 1 \right)} \right)\]
\[ \Rightarrow I = \dfrac{\pi }{8} - \dfrac{1}{4}\]
\[ \Rightarrow I = \dfrac{{\pi - 2}}{8}\]
But this solution is not present in the given options.