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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,
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Verified Answer
The correct answer is:
I $_1 < \mathrm{I}_2$
Hints : $I_1=\int_0^{\pi / 4} \sin ^2 x d x ; I_2=\int_0^{\pi / 4} \cos ^2 x d x$
$$
\begin{aligned}
& \text { In }\left(0, \frac{\pi}{4}\right), \cos ^2 x>\sin ^2 x \therefore \int_0^{\pi / 4} \cos ^2 x d x>\int_0^{\pi / 4} \sin ^2 x d x \\
& I_2>I_1 \text { i.e. } I_1 < I_2
\end{aligned}
$$
$$
\begin{aligned}
& \text { In }\left(0, \frac{\pi}{4}\right), \cos ^2 x>\sin ^2 x \therefore \int_0^{\pi / 4} \cos ^2 x d x>\int_0^{\pi / 4} \sin ^2 x d x \\
& I_2>I_1 \text { i.e. } I_1 < I_2
\end{aligned}
$$
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