Question

If the value of the definite integral $\underset{0}{\overset{2\pi }{\int }}\frac{dx}{2+\sin 2x}=k\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{dx}{3+{\cos }^{2}x}$ then the value of $k$ equals

• A. 16
• B. 8
• C. 4
• D. 2

Answer 1

Answer: (B)

$I=\underset{0}{\overset{2\pi }{\int }}\frac{dx}{2+\sin 2x}=2\underset{0}{\overset{\pi }{\int }}\frac{dx}{2+\sin 2x}$....(1) (As period of function $2+\sin 2x=\pi$ )
Also, $I=2\underset{0}{\overset{\pi }{\int }}\frac{dx}{2-\sin 2x}$...(2)
(Using ${\int }_a^{b}f(x)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$ )
$\therefore$ On adding (1) and (2), we get
$2I=2\underset{0}{\overset{\pi }{\int }}\frac{4dx}{4-{\sin }^{2}2x}$ (Put $2x=t\Rightarrow dx=\frac{dt}{2}$ )
$2I=4\underset{0}{\overset{2\pi }{\int }}\frac{dt}{4-{\sin }^{2}t}=8\underset{0}{\overset{\pi }{\int }}\frac{dt}{4-{\sin }^{2}t}=16\underset{0}{\overset{\pi /2}{\int }}\frac{dt}{4-{\sin }^{2}t}$
$\Rightarrow I=8\underset{0}{\overset{\pi /2}{\int }}\frac{dt}{4-{\sin }^{2}t}=8\underset{0}{\overset{\pi /2}{\int }}\frac{dx}{3+{\cos }^{2}x}$
Hence $k=8$