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Definite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left[\frac{x+\frac{\pi}{4}}{2-\cos 2 x}\right] d x=$
  • A
    $\frac{8 \pi \sqrt{3}}{5}$
  • B
    $\frac{2 \pi \sqrt{3}}{9}$
  • C
    $\frac{4 \pi^2 \sqrt{3}}{9}$
  • ✓
    $\frac{\pi^2}{6 \sqrt{3}}$
✓

Answer

Correct option: D.
$\frac{\pi^2}{6 \sqrt{3}}$
(D)
$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x$
$=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x}{2-\cos 2 x} d x+\frac{\pi}{4} \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{2-\cos 2 x}$
$=0+2 \cdot \frac{\pi}{4} \int_0^{\frac{\pi}{4}} \frac{d x}{2-\cos 2 x}$
$\ldots\left[\begin{array}{l}\because \frac{x}{2-\cos 2 x} \text { is an odd function } \\ \text { and } \frac{1}{2-\cos 2 x} \text { is an even function }\end{array}\right]$
$=\frac{\pi}{2} \int_0^{\frac{\pi}{4}} \frac{d x}{2-\cos 2 x}$
$=\frac{\pi}{2} \int_0^1 \frac{1}{2-\left(\frac{1-t^2}{1+t^2}\right)} \cdot \frac{d t}{1+t^2}$ $\ldots .[$ Put $\tan x=t]$
$\begin{array}{l}=\frac{\pi}{2} \int_0^1 \frac{ dt }{1+3 t ^2}=\frac{\pi}{2} \cdot \frac{1}{\sqrt{3}}\left[\tan ^{-1}(\sqrt{3} t )\right]_0^1 \\ =\frac{\pi}{2 \sqrt{3}}\left[\tan ^{-1}(\sqrt{3})-0\right]=\frac{\pi}{2 \sqrt{3}} \cdot \frac{\pi}{3}=\frac{\pi^2}{6 \sqrt{3}}\end{array}$

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