If $\mathrm{I}_1=\int_0^{\pi / 4} \sin ^2 \mathrm{x} d \mathrm{x}$ and $\mathrm{I}_2=\int_0^{\pi / 4} \cos ^2 \mathrm{x} d \mathrm{x}$, then,

• A

  $\mathrm{I}_1=\mathrm{I}_2$
• Correct answer
  B

  I $_1 < \mathrm{I}_2$
• C

  $\mathrm{I}_1>\mathrm{I}_2$
• D

  $\mathrm{I}_2=\mathrm{I}_1+\pi / 4$

Solution

Hints : $I_1=\int_0^{\pi / 4} \sin ^2 x d x ; I_2=\int_0^{\pi / 4} \cos ^2 x d x$ $$ \begin{aligned} & \text { In }\left(0, \frac{\pi}{4}\right), \cos ^2 x>\sin ^2 x \therefore \int_0^{\pi / 4} \cos ^2 x d x>\int_0^{\pi / 4} \sin ^2 x d x \\ & I_2>I_1 \text { i.e. } I_1 < I_2 \end{aligned} $$

Hence, the correct answer is option: 'B'.