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Q.
If the value of the definite integral $\int\limits_0^{2 \pi} \frac{ dx }{2+\sin 2 x }= k \int\limits_0^{\frac{\pi}{2}} \frac{ dx }{3+\cos ^2 x }$ then the value of $k$ equals
Integrals
Solution:
$I=\int\limits_0^{2 \pi} \frac{d x}{2+\sin 2 x}=2 \int\limits_0^\pi \frac{d x}{2+\sin 2 x}$....(1) (As period of function $2+\sin 2 x=\pi$ )
Also, $I=2 \int\limits_0^\pi \frac{d x}{2-\sin 2 x}$...(2)
(Using $\int_a^b f(x) d x=\int\limits_a^b f(a+b-x) d x$ )
$\therefore $ On adding (1) and (2), we get
$2 I =2 \int\limits_0^\pi \frac{4 dx }{4-\sin ^2 2 x } $ (Put $2 x = t \Rightarrow dx =\frac{ dt }{2}$ )
$2 I =4 \int\limits_0^{2 \pi} \frac{ dt }{4-\sin ^2 t }=8 \int\limits_0^\pi \frac{ dt }{4-\sin ^2 t }=16 \int\limits_0^{\pi / 2} \frac{ dt }{4-\sin ^2 t }$
$\Rightarrow I =8 \int\limits_0^{\pi / 2} \frac{ dt }{4-\sin ^2 t }=8 \int\limits_0^{\pi / 2} \frac{ dx }{3+\cos ^2 x }$
Hence $k = 8$