MCQ
$\int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x=..............$.
- A$\frac{\pi^2}{6}$
- B$\frac{\pi^2}{12 \sqrt{3}}$
- C$\frac{\pi^2}{3 \sqrt{3}}$
- ✓$\frac{\pi^2}{6 \sqrt{3}}$
$x \rightarrow-x$
$I=\int \limits_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{-x+\frac{\pi}{4}}{2-\cos 2 x} d x$
$(1)$ + $(2)$
$2 I=\int \limits_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{\frac{\pi}{2}}{2-\cos 2 x} d x$
$I =\frac{\pi}{4} \cdot 2 \int \limits_0^{\frac{\pi}{4}} \frac{ dx }{2-\cos 2 x } dx$
$I=\frac{\pi}{4} \cdot 2 \int \limits_0^{\frac{\pi}{4}} \frac{\left(1+\tan ^2 x \right) dx }{2\left(1+\tan ^2 x \right)-\left(1-\tan ^2 x \right)}$
$I =\frac{\pi}{4} \int \limits_0^1 \frac{ dt }{3 t ^2+1}$
$\Rightarrow I =\frac{\pi}{2 \sqrt{3}} \tan ^{-1} \sqrt{3}$
$I =\frac{\pi^2}{6 \sqrt{3}}$
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($1$) The value of $\frac{625}{4} p _1$ is
($2$) The value of $\frac{125}{4} p _2$ is
Give the answer or queution ($1$) and ($2$)