`I=int_(0)^(pi)(x)/(1+sinx)dx` पर विचार कीजिए ।
I किसके बराबर है ?

Options

A. `pi`
B. 0
C. `pi`
D. `2pi`

Answer

C

Explanation

दिया है, `I=int_(0)^(pi)(xdx)/((1+sinx))` . . . (i)
`rArrI=int_(0)^(pi)((pi-x))/(1+sin(pi-x))dx`
`[becauseint_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx]`
`rArrI=int_(0)^(pi)((pi-x))/(1+sinx)dx` . . . (ii)
समी (i) व (ii) को जोड़ने पर,
`2I=piint_(0)^(pi)(dx)/(1+sinx)` . . . (iii)
`rArr2I=2piint_(0)^(pi//2)(dx)/(1+sinx)`
`[becauseint_(0)^(2a)f(x)dx=2int_(0)^(a)f(x)dx"यदि"f(2a-x)=f(x)]`
`rArrI=piint_(0)^(pi//2)(dx)/(1+(2tanx//2)/(1+tan^(2)x//2))`
`rArrI=piint_(0)^(pi//2)(sec^(2)x//2dx)/(tan^(2)x//2+1+2tanx//2)`
`rArrI=piint_(0)^(pi//2)(sec^(2)x//2dx)/((tanx//2+1)^(2))`
अब ,`"tan"(x)/(2)+1=t` रखने पर,
`"sec"^(2)(x)/(2)*(1)/(2)dx=dt`
`rArr"sec"^(2)(x)/(2)*dx=2dt`
जब x=0 , तब t=1
जब `x=(pi)/(2)`, तब t=2
`thereforeI=2piint_(1)^(2)(dt)/(d^(2))=-2pi[(1)/(t)]_(1)^(2)`
`=-2pi[(1)/(2)-1]=-2pixx(-(1)/(2))`
`=pi` . . . (iv)