WBJEE · Maths · Definite Integration
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 \(\mathrm{I}_1=\mathrm{I}_2\)
- B I \(_1 < \mathrm{I}_2\)
- C \(\mathrm{I}_1>\mathrm{I}_2\)
- D \(\mathrm{I}_2=\mathrm{I}_1+\pi / 4\)
Answer & Solution
Correct Answer
(B) I \(_1 < \mathrm{I}_2\)
Step-by-step Solution
Detailed explanation
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
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