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
Integrate the functions:
tan2(2x – 3)
Integrate the functions:
`cos sqrt(x)/sqrtx`
Integrate the functions:
`((x+1)(x + logx)^2)/x`
Evaluate: `int_0^3 f(x)dx` where f(x) = `{(cos 2x, 0<= x <= pi/2),(3, pi/2 <= x <= 3) :}`
Write a value of\[\int \cos^4 x \text{ sin x dx }\]
Write a value of\[\int\frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} dx\]
Write a value of
The value of \[\int\frac{1}{x + x \log x} dx\] is
Integrate the following w.r.t. x : `(3x^3 - 2x + 5)/(xsqrt(x)`
Evaluate the following integrals : `int cos^2x.dx`
Integrate the following functions w.r.t. x : `(e^(2x) + 1)/(e^(2x) - 1)`
Integrate the following functions w.r.t.x:
`(5 - 3x)(2 - 3x)^(-1/2)`
Integrate the following functions w.r.t. x : `(4e^x - 25)/(2e^x - 5)`
Evaluate the following : `int (1)/(4 + 3cos^2x).dx`
Integrate the following functions w.r.t. x : `int (1)/(3 + 2 sin2x + 4cos 2x).dx`
Evaluate the following integrals:
`int (7x + 3)/sqrt(3 + 2x - x^2).dx`
Choose the correct option from the given alternatives :
`int (1 + x + sqrt(x + x^2))/(sqrt(x) + sqrt(1 + x))*dx` =
Choose the correct options from the given alternatives :
`int sqrt(cotx)/(sinx*cosx)*dx` =
Evaluate `int 1/(x (x - 1))` dx
Fill in the Blank.
`int (5("x"^6 + 1))/("x"^2 + 1)` dx = x4 + ______ x3 + 5x + c
Evaluate `int 1/((2"x" + 3))` dx
`int(1 - x)^(-2) dx` = ______.
Evaluate `int"e"^x (1/x - 1/x^2) "d"x`
`int (x^2 + 1)/(x^4 - x^2 + 1)`dx = ?
`int ("e"^x(x + 1))/(sin^2(x"e"^x)) "d"x` = ______.
Evaluate the following.
`int(20 - 12"e"^"x")/(3"e"^"x" - 4) "dx"`
`int "cosec"^4x dx` = ______.
Evaluate the following:
`int x^3/(sqrt(1+x^4))dx`
Evaluate `int(1 + x + x^2 / (2!))dx`
Evaluate `int 1/(x(x-1)) dx`
Evaluate:
`intsqrt(sec x/2 - 1)dx`
Which standard substitution is used for \[\sqrt{\mathrm{a}^2-x^2}\], \[\frac{1}{\sqrt{\mathrm{a}^2-x^2}}\], or \[\mathrm{a}^2-x^2\]?
Which standard substitution is used for \[\sqrt{x^2+\mathrm{a}^2}\], \[\frac{1}{\sqrt{x^2+\mathrm{a}^2}}\], or \[x^2+\mathrm{a}^2\]?
After choosing \[u=g(x)\], what is the next step?
After putting \[t=\cos x\], which integral is obtained from \[\int\sin^2x\cos^2x(\sin x)\,dx\]?
