Advertisements
Advertisements
प्रश्न
Integrate the following functions w.r.t. x : `(cos3x - cos4x)/(sin3x + sin4x)`
उत्तर
Let I = `int(cos3x - cos4x)/(sin3x + sin4x).dx`
= `int(-2sin((3x + 4x)/2)sin((3x - 4x)/2))/(2sin((3x + 4x)/2)cos((3x - 4x)/2)).dx`
= `int - sin(-x/2)/cos(-x/2).dx`
= `int sin(x/2)/cos(x/2).dx`
= `int tan(x/2).dx`
= `log|sec (x/2)|/((1/2)) + c`
= `2log|sec (x/2)| + c`.
APPEARS IN
संबंधित प्रश्न
Evaluate : `int_0^pi(x)/(a^2cos^2x+b^2sin^2x)dx`
Find `intsqrtx/sqrt(a^3-x^3)dx`
Evaluate :
`∫(x+2)/sqrt(x^2+5x+6)dx`
Integrate the functions:
sin x ⋅ sin (cos x)
Integrate the functions:
`x/(sqrt(x+ 4))`, x > 0
Integrate the functions:
`x/(e^(x^2))`
Integrate the functions:
`1/(cos^2 x(1-tan x)^2`
Integrate the functions:
`1/(1 + cot x)`
Evaluate : `∫1/(3+2sinx+cosx)dx`
Evaluate `int 1/(3+ 2 sinx + cosx) dx`
Write a value of\[\int\text{ tan x }\sec^3 x\ dx\]
Write a value of\[\int\frac{1}{1 + 2 e^x} \text{ dx }\].
Write a value of\[\int e^{ax} \left\{ a f\left( x \right) + f'\left( x \right) \right\} dx\] .
Write a value of\[\int\sqrt{4 - x^2} \text{ dx }\]
Evaluate: \[\int\frac{x^3 - 1}{x^2} \text{ dx}\]
Evaluate : `int ("e"^"x" (1 + "x"))/("cos"^2("x""e"^"x"))"dx"`
Integrate the following w.r.t. x:
`3 sec^2x - 4/x + 1/(xsqrt(x)) - 7`
Evaluate the following integrals : `int(5x + 2)/(3x - 4).dx`
Evaluate the following integrals:
`int (sin4x)/(cos2x).dx`
Integrate the following function w.r.t. x:
x9.sec2(x10)
Integrate the following functions w.r.t. x : `(1)/(sqrt(x) + sqrt(x^3)`
Evaluate the following : `int (1)/(4x^2 - 3).dx`
Evaluate the following : `int sqrt((9 + x)/(9 - x)).dx`
Evaluate the following : `int (1)/(x^2 + 8x + 12).dx`
Evaluate the following : `(1)/(4x^2 - 20x + 17)`
Evaluate the following : `int (1)/sqrt(x^2 + 8x - 20).dx`
Choose the correct options from the given alternatives :
`int sqrt(cotx)/(sinx*cosx)*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 (e^(2x) + e^-2x)/e^x*dx` =
Evaluate `int (-2)/(sqrt("5x" - 4) - sqrt("5x" - 2))`dx
Evaluate `int (3"x"^2 - 5)^2` dx
Evaluate the following.
`int (1 + "x")/("x" + "e"^"-x")` dx
Evaluate the following.
`int 1/("x"^2 + 4"x" - 5)` dx
Evaluate the following.
`int 1/("a"^2 - "b"^2 "x"^2)` dx
`int (x^2 + x - 6)/((x - 2)(x - 1))dx = x` + ______ + c
Evaluate `int (5"x" + 1)^(4/9)` dx
Evaluate: If f '(x) = `sqrt"x"` and f(1) = 2, then find the value of f(x).
Evaluate: ∫ |x| dx if x < 0
Evaluate: `int 1/(2"x" + 3"x" log"x")` dx
Evaluate: `int "e"^"x" (1 + "x")/(2 + "x")^2` dx
`int sqrt(x^2 + 2x + 5)` dx = ______________
`int cos sqrtx` dx = _____________
If `int 1/(x + x^5)` dx = f(x) + c, then `int x^4/(x + x^5)`dx = ______
`int (sin4x)/(cos 2x) "d"x`
`int ("e"^(3x))/("e"^(3x) + 1) "d"x`
`int ("e"^(2x) + "e"^(-2x))/("e"^x) "d"x`
Choose the correct alternative:
`int(1 - x)^(-2) dx` = ______.
To find the value of `int ((1 + logx))/x` dx the proper substitution is ______
`int sin^-1 x`dx = ?
`int (sin (5x)/2)/(sin x/2)dx` is equal to ______. (where C is a constant of integration).
`int(3x + 1)/(2x^2 - 2x + 3)dx` equals ______.
If `int sinx/(sin^3x + cos^3x)dx = α log_e |1 + tan x| + β log_e |1 - tan x + tan^2x| + γ tan^-1 ((2tanx - 1)/sqrt(3)) + C`, when C is constant of integration, then the value of 18(α + β + γ2) is ______.
Find `int dx/sqrt(sin^3x cos(x - α))`.
Evaluate `int_(logsqrt(2))^(logsqrt(3)) 1/((e^x + e^-x)(e^x - e^-x)) dx`.
`int secx/(secx - tanx)dx` equals ______.
If f′(x) = 4x3 − 3x2 + 2x + k, f(0) = -1 and f(1) = 4, find f(x)
Prove that:
`int 1/sqrt(x^2 - a^2) dx = log |x + sqrt(x^2 - a^2)| + c`.
Evaluate `int (1)/(x(x - 1))dx`
Evaluate `int 1/(x(x-1))dx`
`int x^2/sqrt(1 - x^6)dx` = ______.
Evaluate the following.
`intx sqrt(1 +x^2) dx`
Evaluate 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:
`int(5x^2-6x+3)/(2x-3)dx`
Evaluate the following.
`intx^3/sqrt(1+x^4)dx`
Evaluate the following.
`int1/(x^2 + 4x - 5)dx`
Evaluate:
`intsqrt(sec x/2 - 1)dx`
