समाकलनों`I_(1)=int_(pi//6)^(pi//3)(dx)/(1+sqrt(tanx))`
और `I_(2)=int_(pi//6)^(pi//3)(sqrt(sinx)dx)/(sqrt(sinx)+sqrt(cosx))` पर विचार कीजिए।
`I_(1)-I_(2)` किसके तुल्य है ?

Options

A. 0
B. `2//_(1)`
C. `pi`
D. इनमें से कोई नहीं

Answer

A

Explanation

दिया है, `I_(1)=int_(pi//6)^(pi//3)(dx)/(1+sqrt(tanx))`
तथा `I_(2)=int_(pi//6)^(pi//3)(sqrt(sinx))/(sqrt(sinx)+sqrt(cosx))dx` . . . (i)
`I_(1)` को निम्न प्रकार लिखा जा सकता है
`I_(1)=int_(pi//6)^(pi//3)(dx)/(1+(sqrt(sinx))/(sqrt(cosx)))`
`=int_(pi//6)^(pi//3)(sqrt(cosx))/(sqrt(sinx)+sqrt(cosx))dx`. . . (ii)
अब,
`I_(1)-I_(2)=int_(pi//6)^(pi//3)(sqrt(cosx))/(sqrt(sinx)+sqrt(cosx))dx`
`-int_(pi//6)^(pi//3)(sqrt(sinx))/(sqrt(sinx)+sqrt(cosx))dx`
`=int_(pi//6)^(pi//3)((sqrt(cosx)-sqrt(sinx))/(sqrt(cosx)+sqrt(sinx)))dx`
`=int_(pi//6)^(pi//3)`
`(sqrtcos((pi)/(3)+(pi)/(6)-x)-sqrt(sin((pi)/(3)+(pi)/(6)-x)))/(sqrt(cos((pi)/(3)+(pi)/(6)-x))+sqrt(sin((pi)/(3)+(pi)/(6)-x)))dx`
`[becauseint_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx]`
`=int_(pi//6)^(pi//3)(sqrt(cos((pi)/(2)-x))-sqrt(sin((pi)/(2)-x)))/(sqrt(cos((pi)/(2)-x))+sqrt(sin((pi)/(2)x)))dx`
`=int_(pi//6)^(pi//3)(sqrt(sinx)-sqrt(cosx))/(sqrt(sinx)+sqrt(cosx))dx`
`=-{int_(pi//6)^(pi//3)(sqrt(cosx)-sqrt(sinx))/(sqrt(sinx)+sqrt(cosx))dx}`
`=-(I_(1)-I_(2)rArr2(I_(1)-I_(2))=0`
`rArrI_(1)-I_(2)=0`