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प्रश्न
Prove that: `int "dx"/(sqrt("x"^2 +"a"^2)) = log |"x" +sqrt("x"^2 +"a"^2) | + "c"`
उत्तर
Let I = `int 1/sqrt("x"^2 + "a"^2) "dx"`
Put x = a tan θ ⇒ tan θ = `"x"/"a"`
∴ dx = a sec2 θ dθ
∴ I = `int 1/ sqrt("a"^2 "tan"^2 theta +"a"^2) "a" "sec"^2 theta "d" theta`
= `int ("a"."sec"^2 theta)/("a" sqrt(1+"tan"^2 theta)) "d"theta`
= `int ("sec"^2 theta)/("sec" theta) "d"theta `
`= int "sec" theta . "d" theta`
`= "log" |"sec" theta +"tan" theta| +"c"_1`
`= "log" |"x"/"a" + sqrt("sec"^2 theta)| + "c"_1`
`= "log" | "x"/"a" + sqrt 1+ "tan"^2 theta | + "c"_1`
=`"log" |"x" /"a" +sqrt(1+"x"^2/"a"^2)| +"c"_1`
=` "log" |"x"/"a" + sqrt( "a"^2 + "x"^2)/"a"| + "c"_1`
`= "log" |"x" +sqrt("x"^2 +"a"^2)| - "log" "a" + "c"_1`
`therefore int 1/sqrt("x"^2 + "a"^2) "dx" = "log" |"x" +sqrt("x"^2 +"a"^2)| - "log" "a" + "c" ,`
where c = - log a +c1
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