AP EAMCET20225 Jul 2022Evening ShiftMathematicsIndefinite IntegrationActual
There exist such that a >| | , then d x 1+a x =
Options
- A1 a^2-1 ⁻¹ ( a-1 a+1 x 2 )+C
- B1 a^2-1 ⁻¹ ( 1-a 1+a x 2 )+C
- C1 a^2-1 ( a+1 x 2 - a-1 x 2 a-1 x 2 + a-1 x 2 )+C
- D1 a^2-1 ( a+1 x 2 + a-1 x 2 a+1 x 2 - a-1 x 2 )+C
Correct answer
D. 1 a^2-1 ( a+1 x 2 + a-1 x 2 a+1 x 2 - a-1 x 2 )+C
Step-by-step solution
Given, 1 1+a (x) d x We are substituting (x)= 1- ^2 ( x 2 ) 1+ ^2 ( x 2 ) . 1 1+a (x) d x= 1 1+ a (1- ^2 ( x 2 ) ) 1+ ^2 ( x 2 ) d x Multiplying both numerator and denominator by 1+ ^2 ( x 2 )= 1+ ^2 ( x 2 ) (1+a)+(1-a) ^2 ( x 2 ) d x= ^2 ( x 2 ) (1+a)+(1-a) ^2 ( x 2 ) d x (Using the trigonometric formula ^2(x)- ^2(x)=1 ). Putting ( x 2 )=z Tanking derivative both sides, d ( ( x 2 ) )=d z ^2 ( x 2 ) d x=2 d z (Using the formula . d( (x)) d x = ^2(x) ) . Now, replacing x in the terms of z . = 1 (1+a)+(1-a) z^2 2 d z