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REPEATER 2024 — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsREPEATER 20242 Marks
Question
Evaluate :
$\underset{\overset{\int}{0}}{{\frac{\pi}{4}}} \sqrt{1+\sin 2 x} \cdot d x$
✓

Answer

Let$\quad$$I=\underset{\overset{\int}{0}}{{\frac{\pi}{4}}} \sqrt{1+\sin 2 x} \cdot d x$
$\therefore \quad=\underset{\overset{\int}{0}}{{\frac{\pi}{4}}} \sqrt{(\cos x+\sin x)^2} \cdot d x \quad \ldots \ldots \ldots \ldots .\left[\because 1+\sin 2 \theta=(\cos \theta+\sin \theta)^2\right]$
$\therefore \quad=\underset{\overset{\int}{0}}{{\frac{\pi}{4}}}(\cos x+\sin x) \cdot d x$
$\therefore \quad=[\sin x-\cos x]_0^{\frac{\pi}{4}}$
$\therefore \quad=\left[\left(\sin \frac{\pi}{4}-\cos \frac{\pi}{4}\right)-(\sin 0-\cos 0)\right]$
$\therefore \quad=\left[\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)-(0-1)\right]$
$\begin{array}{ll}\therefore & =1\end{array}$
$\therefore \quad \underset{\overset{\int}{0}}{{\frac{\pi}{4}}} \sqrt{1+\sin 2 x} \cdot d x=1$

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