Advertisements
Advertisements
प्रश्न
Prove that: `int "dx"/(sqrt("x"^2 +"a"^2)) = log |"x" +sqrt("x"^2 +"a"^2) | + "c"`
Advertisements
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
APPEARS IN
संबंधित प्रश्न
Find: `int(x+3)sqrt(3-4x-x^2dx)`
Find `intsqrtx/sqrt(a^3-x^3)dx`
Integrate the functions:
`1/(x-sqrtx)`
Integrate the functions:
`e^(2x+3)`
Write a value of
Write a value of\[\int\left( e^{x \log_e \text{ a}} + e^{a \log_e x} \right) dx\] .
Write a value of\[\int\sqrt{9 + x^2} \text{ dx }\].
Evaluate the following integrals : `intsqrt(1 - cos 2x)dx`
Evaluate the following integrals: `int (2x - 7)/sqrt(4x - 1).dx`
Evaluate the following integrals:
`int (sin4x)/(cos2x).dx`
Integrate the following functions w.r.t. x : `(logx)^n/x`
Integrate the following functions w.r.t. x : `(e^(2x) + 1)/(e^(2x) - 1)`
Integrate the following functions w.r.t. x : e3logx(x4 + 1)–1
Integrate the following functions w.r.t.x:
`(5 - 3x)(2 - 3x)^(-1/2)`
Integrate the following functions w.r.t. x : `(sinx + 2cosx)/(3sinx + 4cosx)`
Evaluate the following : `int (1)/(1 + x - x^2).dx`
Evaluate the following : `(1)/(4x^2 - 20x + 17)`
Evaluate the following : `int (1)/sqrt(8 - 3x + 2x^2).dx`
Evaluate the following integrals : `int (3x + 4)/sqrt(2x^2 + 2x + 1).dx`
Choose the correct options from the given alternatives :
`2 int (cos^2x - sin^2x)/(cos^2x + sin^2x)*dx` =
Choose the correct options from the given alternatives :
`int (cos2x - 1)/(cos2x + 1)*dx` =
Evaluate the following.
`int 1/(sqrt"x" + "x")` dx
Evaluate:
`int (5x^2 - 6x + 3)/(2x − 3)` dx
`int 1/(xsin^2(logx)) "d"x`
`int (7x + 9)^13 "d"x` ______ + c
Evaluate `int(3x^2 - 5)^2 "d"x`
`int x^3"e"^(x^2) "d"x`
`int ((x + 1)(x + log x))^4/(3x) "dx" =`______.
`int ("d"x)/(x(x^4 + 1))` = ______.
`int ("e"^x(x + 1))/(sin^2(x"e"^x)) "d"x` = ______.
Evaluate `int(1 + x + x^2/(2!) )dx`
Evaluated the following
`int x^3/ sqrt (1 + x^4 )dx`
Evaluate the following.
`int(1)/(x^2 + 4x - 5)dx`
Evaluate the following.
`int x^3/sqrt(1+x^4) dx`
Evaluate:
`intsqrt(sec x/2 - 1)dx`
`int (x + 1)/(x(1 + xe^x)) dx` is equal to
Which substitution is appropriate for \[\sqrt{\frac{x}{a-x}}\], \[\sqrt{\frac{a-x}{x}}\], \[\sqrt{x(a-x)}\], or \[\frac{1}{\sqrt{x(a-x)}}\]?
After choosing \[u=g(x)\], what is the next step?
