Advertisements
Advertisements
प्रश्न
`int sqrt(tanx) + sqrt(cotx) "d"x`
Advertisements
उत्तर
Let I = `int (sqrt(tanx) + sqrt(cotx)) "d"x`
= `int (sqrt(tanx) + 1/sqrt(tanx)) "d"x`
= `int (tanx + 1)/sqrt(tanx) "d"x`
Put `sqrt(tanx)` = t
∴ tanx = t2
∴x = tan−1(t2)
∴ dx = `1/(1 + ("t"^2)^2) * 2"t" "dt"`
∴ dx = `(2"t")/(1 + "t"^4) "dt"`
∴ I = `int ("t"^2 + 1)/"t"* (2"t")/(1 + "t"^4) "dt"`
= `2 int ("t"^2 + 1)/("t"^4 + 1) "dt"`
= `2 int (1 + 1/"t"^2)/("t"^2 + 1/"t"^2) "dt"`
= `2 int (1 + 1/"t"^2)/(("t" - 1/"t")^2 + 2)`
Put `"t" - 1/"t"` = u
∴ `(1 + 1/"t"^2) "dt"` = du
∴ I = `2 int "du"/("u"^2 + 2)`
= `2 int "du"/("u"^2 + (sqrt(2))^2`
= `2* 1/sqrt(2)tan^-1 ("u"/sqrt(2)) + "c"`
= `sqrt(2)tan^-1 (("t" - 1/"t")/sqrt(2)) + "c"`
= `sqrt(2)tan^-1 (("t"^2 - 1)/sqrt(2)) + "c"`
= `sqrt(2)tan^-1 ((tanx - 1)/sqrt(2tanx)) + "c"`
APPEARS IN
संबंधित प्रश्न
Prove that: `int sqrt(a^2 - x^2) * dx = x/2 * sqrt(a^2 - x^2) + a^2/2 * sin^-1(x/a) + c`
Integrate : sec3 x w. r. t. x.
Prove that:
`int sqrt(x^2 - a^2)dx = x/2sqrt(x^2 - a^2) - a^2/2log|x + sqrt(x^2 - a^2)| + c`
If u and v are two functions of x then prove that
`intuvdx=uintvdx-int[du/dxintvdx]dx`
Hence evaluate, `int xe^xdx`
Integrate the function in x sec2 x.
`int e^x sec x (1 + tan x) dx` equals:
Evaluate the following : `int cos sqrt(x).dx`
Evaluate the following : `int sin θ.log (cos θ).dθ`
Evaluate the following : `int(sin(logx)^2)/x.log.x.dx`
Evaluate the following: `int logx/x.dx`
Evaluate the following:
`int x.sin 2x. cos 5x.dx`
Integrate the following functions w.r.t. x : `sqrt(4^x(4^x + 4))`
Integrate the following functions w.r.t. x : `e^x .(1/x - 1/x^2)`
Integrate the following functions w.r.t.x:
`e^(5x).[(5x.logx + 1)/x]`
Choose the correct options from the given alternatives :
`int (sin^m x)/(cos^(m+2)x)*dx` =
Integrate the following with respect to the respective variable : `(sin^6θ + cos^6θ)/(sin^2θ*cos^2θ)`
Integrate the following w.r.t. x: `(1 + log x)^2/x`
Integrate the following w.r.t.x : `sqrt(x)sec(x^(3/2))*tan(x^(3/2))`
Integrate the following w.r.t.x : log (x2 + 1)
Evaluate the following.
`int x^2 *e^(3x)`dx
Evaluate the following.
`int "e"^"x" "x"/("x + 1")^2` dx
`int ["cosec"(logx)][1 - cot(logx)] "d"x`
`int 1/(x^2 - "a"^2) "d"x` = ______ + c
Evaluate `int (2x + 1)/((x + 1)(x - 2)) "d"x`
`int "e"^x x/(x + 1)^2 "d"x`
`int logx/(1 + logx)^2 "d"x`
`int log x * [log ("e"x)]^-2` dx = ?
`int "e"^x int [(2 - sin 2x)/(1 - cos 2x)]`dx = ______.
Evaluate the following:
`int_0^pi x log sin x "d"x`
`int tan^-1 sqrt(x) "d"x` is equal to ______.
Find: `int e^x.sin2xdx`
`int(logx)^2dx` equals ______.
Evaluate:
`int(1+logx)/(x(3+logx)(2+3logx)) dx`
Evaluate `int(3x-2)/((x+1)^2(x+3)) dx`
The integrating factor of `ylogy.dx/dy+x-logy=0` is ______.
`int logx dx = x(1+logx)+c`
Evaluate the following.
`int (x^3)/(sqrt(1 + x^4))dx`
Evaluate:
`int e^(logcosx)dx`
If u and v are two differentiable functions of x, then prove that `intu*v*dx = u*intv dx - int(d/dx u)(intv dx)dx`. Hence evaluate: `intx cos x dx`
Complete the following activity:
`int_0^2 dx/(4 + x - x^2) `
= `int_0^2 dx/(-x^2 + square + square)`
= `int_0^2 dx/(-x^2 + x + 1/4 - square + 4)`
= `int_0^2 dx/ ((x- 1/2)^2 - (square)^2)`
= `1/sqrt17 log((20 + 4sqrt17)/(20 - 4sqrt17))`
Evaluate the following:
`intx^3e^(x^2)dx`
Evaluate the following.
`intx^3/sqrt(1+x^4)dx`
If f'(x) = 4x3 - 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x)
The value of `inta^x.e^x dx` equals
Evaluate:
`inte^x "cosec" x(1 - cot x)dx`
