Definite Integration Question 23
Question 23 - 01 February - Shift 2
The value of the integral $\int _{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x$ is
(1) $\frac{\pi^{2}}{6}$
(2) $\frac{\pi^{2}}{12 \sqrt{3}}$
(3) $\frac{\pi^{2}}{3 \sqrt{3}}$
(4) $\frac{\pi^{2}}{6 \sqrt{3}}$
Show Answer
Answer: (4)
Solution:
Formula: Integration by substitution, Double Angle Identities, Properties of definite integral
$ I=\int _{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x \ldots (1) $
Replace $x$ by $x$
$ I=\int _{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{-x+\frac{\pi}{4}}{2-\cos 2 x} d x \ldots (2) $
On adding $(1) $ and $ (2)$
$ 2 I=\int _{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{\frac{\pi}{2}}{2-\cos 2 x} d x $
$I=\frac{\pi}{4} \cdot 2 \int_0^{\frac{\pi}{4}} \frac{dx}{2-\cos 2 x} dx, \quad$ Using definite integral property
$ \begin{aligned} & I=\frac{\pi}{4} \cdot 2 \int_0^{\frac{\pi}{4}} \frac{(1+\tan ^{2} x) dx}{2(1+\tan ^{2} x)-(1-\tan ^{2} x)} \\ &I=\frac{\pi}{4} \cdot 2 \int_0^{\frac{\pi}{4}} \frac{(\sec ^{2} x) dx}{3\tan ^{2} x+1}\\ & I=\frac{\pi}{4} \int_0^{1} \frac{dt \text{}}{3 t^{2}+1} \quad \text{substituting} \quad tanx =t\\ & \Rightarrow I=\frac{\pi}{2 \sqrt{3}} \tan ^{-1} \sqrt{3} \\ & I=\frac{\pi^{2}}{6 \sqrt{3}} \end{aligned} $
