Question

निम्नलिखित के मान निकालिए-

`int sqrt(("a" + x)/("a" - x)) "d"x`

Answer 1

मान लीजिए I = `int sqrt(("a" + x)/("a" - x)) "d"x`

x = `"a" cos 2theta` रखिए

⇒ dx = `-"a" * sin 2theta * 2 * "d"theta`

∴ I = `-2int sqrt(("a" + "a" cos 2theta)/("a" -  "a" cos 2theta)) * "a" sin 2theta "d"theta`

= `-2"a" int sqrt((1 + cos 2theta)/(1 - cos 2theta)) sin 2theta "d"theta`

= `-2"a" int sqrt((2 cos^2theta)/(2 sin^2 theta)) sin 2theta "d"theta`

= `-2"a" int cot theta * sin 2theta "d"theta`

= `-2"a" int costheta/sintheta * 2 sin theta cos theta "d" theta`

= `-4"a" int cos^2theta "d"theta`

= `-2"a" int (1 + cos 2theta)"d"theta`

= `-2"a" [theta + 1/2 sin 2theta] + "C"`

= `-2"a" [1/2 cos^-1  x/"a" + 1/2 sqrt(1 - x^2/"a"^2)] + "C"`

= `-"a" [cos^-1(x/"a") + sqrt(1 - x^2/"a"^2)] + "C"`