Advertisements
Advertisements
प्रश्न
Integrate the following functions w.r.t. x : `(1)/(2 + 3tanx)`
Advertisements
उत्तर
Let I = `int (1)/(2 + 3tanx).dx`
= `int(1)/(2 + 3(sinx/cosx)).dx`
= `int cosx/(2cosx + 3sinx).dx`
Put,
Numerator = `"A (Denominator) + B"[d/dx("Denominator")]`
∴ cos x = `"A"(2cosx + 3 sinx ) + "B"[d/dx(2cos x + 3 sin x)]`
= A(2 cos x + 3 sin x) + B(– 2 sin x + 3 cos x)
∴ cos x = (2A + 3B)cos x + (3A – 2B)sin x
Equating the coefficients of cos x sin x on both the sides, we get
2A 3B = 1 ...(1)
and
3A – 2B = 0 ...(2)
Multiplying equation (1) by 22 and equation (2) by 3, we get
4A +6B = 2
9A – 6B = 0
On adding, we get
13A = 2
∴ A = `(2)/(13)`
∴ from (2), 2B = 3A = `3(2/13) = (6)/(13)`
∴ B = `(3)/(13)`
∴ cos x = `(2)/(13)(2cosx + 3sinx) + (3)/(13)(-2sinx + 3cosx)`
∴ I = `int[(2/13(2cosx + 3sinx) + 3/13(-2 sinx + 3cosx))/(2cosx + 3sinx)].dx`
= `int[2/13 + (3/13(-2sinx + 3cosx))/(2cosx + 3sinx)].dx`
= `(2)/(13) 1 dx + (3)/(13) int (-2sinx + 3cosx)/(2cosx + 3sinx).dx`
= `(2)/(13)x + (3)/(13)log|2cos x + 3sinx| + c. ...[∵ int (f'(x))/f(x)dx = log|f(x)| + c]`
APPEARS IN
संबंधित प्रश्न
Find: `int(x+3)sqrt(3-4x-x^2dx)`
Integrate the functions:
`sqrt(ax + b)`
Integrate the functions:
`xsqrt(1+ 2x^2)`
Integrate the functions:
`x^2/(2+ 3x^3)^3`
Integrate the functions:
`1/(1 + cot x)`
Integrate the functions:
`1/(1 - tan x)`
Write a value of
Evaluate: \[\int\frac{x^3 - 1}{x^2} \text{ dx}\]
The value of \[\int\frac{\cos \sqrt{x}}{\sqrt{x}} dx\] is
Evaluate the following integrals : `int(5x + 2)/(3x - 4).dx`
Integrate the following functions w.r.t. x : `(1 + x)/(x.sin (x + log x)`
Integrate the following functions w.r.t. x : `(x.sec^2(x^2))/sqrt(tan^3(x^2)`
Integrate the following functions w.r.t. x : `(7 + 4 + 5x^2)/(2x + 3)^(3/2)`
Integrate the following functions w.r.t. x : `(sin6x)/(sin 10x sin 4x)`
Evaluate the following : `int sqrt((9 + x)/(9 - x)).dx`
Evaluate the following : `int (1)/(5 - 4x - 3x^2).dx`
Evaluate the following : `int (1)/sqrt(3x^2 + 5x + 7).dx`
Integrate the following functions w.r.t. x : `int (1)/(2 + cosx - sinx).dx`
Evaluate the following.
`int 1/("x"^2 + 4"x" - 5)` dx
Fill in the Blank.
`int 1/"x"^3 [log "x"^"x"]^2 "dx" = "P" (log "x")^3` + c, then P = _______
Evaluate:
`int (5x^2 - 6x + 3)/(2x − 3)` dx
Evaluate: `int log ("x"^2 + "x")` dx
Evaluate: `int sqrt("x"^2 + 2"x" + 5)` dx
`int (2 + cot x - "cosec"^2x) "e"^x "d"x`
`int sqrt(x) sec(x)^(3/2) tan(x)^(3/2)"d"x`
`int cot^2x "d"x`
`int(1 - x)^(-2) dx` = ______.
State whether the following statement is True or False:
If `int x "f"(x) "d"x = ("f"(x))/2`, then f(x) = `"e"^(x^2)`
`int(5x + 2)/(3x - 4) dx` = ______
`int "e"^(sin^-1 x) ((x + sqrt(1 - x^2))/(sqrt1 - x^2)) "dx" = ?`
General solution of `(x + y)^2 ("d"y)/("d"x) = "a"^2, "a" ≠ 0` is ______. (c is arbitrary constant)
`int[ tan (log x) + sec^2 (log x)] dx= ` ______
If f'(x) = `x + 1/x`, then f(x) is ______.
`int(log(logx) + 1/(logx)^2)dx` = ______.
The integral `int ((1 - 1/sqrt(3))(cosx - sinx))/((1 + 2/sqrt(3) sin2x))dx` is equal to ______.
`int cos^3x dx` = ______.
Evaluate the following.
`int x^3/(sqrt(1+x^4))dx`
Evaluate `int 1/("x"("x" - 1)) "dx"`
Evaluate `int (1)/(x(x - 1))dx`
Evaluate the following.
`int1/(x^2 + 4x - 5) dx`
Evaluate `int(5x^2-6x+3)/(2x-3)dx`
If f'(x) = 4x3 - 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).
If f'(x) = 4x3 – 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).
Which standard substitution is used for \[\sqrt{\mathrm{a}^2-x^2}\], \[\frac{1}{\sqrt{\mathrm{a}^2-x^2}}\], or \[\mathrm{a}^2-x^2\]?
For \[\int\frac{\sin x}{\sin(x+a)}\,dx\], which substitution gives \[dx=dt\]?
