Advertisements
Advertisements
प्रश्न
`int (sinx)/(sin3x) "d"x`
Advertisements
उत्तर
Let I = `int (sinx)/(sin3x) "d"x`
= `int sin x/(3sin x - 4 sin^3x)* "d"x`
= `int sinx/(sinx(3 - 4sin^2x))* "d"x`
= `int 1/(3 - 4sin^2x) "d"x`
Dividing numerator and denominator by cos2x, we get
I = `int (sec^2x)/(3sec^2x - 4tan^2x) * "d"x`
= `int (sec^2x)/(3(1 + tan^2x) - 4tan^2x)* "d"x`
= `int (sec^2x)/(3 - tan^2x) "d"x`
Put tan x = t
∴ sec2x dx = dt
∴ I = `int "dt"/(3 - "t"^2)`
= `int 1/((sqrt(3))^2 - "t"^2) "dt"`
=`1/(2sqrt(3)) log|(sqrt(3) + "t")/(sqrt(3) - "t")| + "c"`
∴ I = `1/(2sqrt(3)) log|(sqrt(3) + tanx)/(sqrt(3) - tanx)| + "c'`
APPEARS IN
संबंधित प्रश्न
Integrate the rational function:
`(2x)/(x^2 + 3x + 2)`
Integrate the rational function:
`(1 - x^2)/(x(1-2x))`
Integrate the rational function:
`(3x + 5)/(x^3 - x^2 - x + 1)`
Integrate the rational function:
`(2x - 3)/((x^2 -1)(2x + 3))`
Integrate the rational function:
`(5x)/((x + 1)(x^2 - 4))`
Integrate the rational function:
`(cos x)/((1-sinx)(2 - sin x))` [Hint: Put sin x = t]
Integrate the rational function:
`(2x)/((x^2 + 1)(x^2 + 3))`
Evaluate : `∫(x+1)/((x+2)(x+3))dx`
Integrate the following w.r.t. x : `(12x + 3)/(6x^2 + 13x - 63)`
Integrate the following w.r.t. x : `(x^2 + x - 1)/(x^2 + x - 6)`
Integrate the following w.r.t. x:
`(6x^3 + 5x^2 - 7)/(3x^2 - 2x - 1)`
Integrate the following w.r.t. x : `(3x - 2)/((x + 1)^2(x + 3)`
Integrate the following w.r.t. x: `(1)/(sinx + sin2x)`
Integrate the following w.r.t. x : `(1)/(2sinx + sin2x)`
Integrate the following w.r.t. x : `(1)/(sinx*(3 + 2cosx)`
Choose the correct options from the given alternatives :
If `int tan^3x*sec^3x*dx = (1/m)sec^mx - (1/n)sec^n x + c, "then" (m, n)` =
Integrate the following with respect to the respective variable : `cot^-1 ((1 + sinx)/cosx)`
Integrate the following w.r.t.x : `(1)/((1 - cos4x)(3 - cot2x)`
Integrate the following w.r.t.x : `sec^2x sqrt(7 + 2 tan x - tan^2 x)`
Evaluate: `int (2"x"^3 - 3"x"^2 - 9"x" + 1)/("2x"^2 - "x" - 10)` dx
Evaluate: `int (1 + log "x")/("x"(3 + log "x")(2 + 3 log "x"))` dx
`int 1/(x(x^3 - 1)) "d"x`
If f'(x) = `x - 3/x^3`, f(1) = `11/2` find f(x)
`int 1/(2 + cosx - sinx) "d"x`
`int "e"^x ((1 + x^2))/(1 + x)^2 "d"x`
`int (6x^3 + 5x^2 - 7)/(3x^2 - 2x - 1) "d"x`
`int ("d"x)/(2 + 3tanx)`
`int (3x + 4)/sqrt(2x^2 + 2x + 1) "d"x`
`int (x + sinx)/(1 - cosx) "d"x`
`int xcos^3x "d"x`
`int (3"e"^(2x) + 5)/(4"e"^(2x) - 5) "d"x`
Choose the correct alternative:
`int (x + 2)/(2x^2 + 6x + 5) "d"x = "p"int (4x + 6)/(2x^2 + 6x + 5) "d"x + 1/2 int 1/(2x^2 + 6x + 5)"d"x`, then p = ?
Choose the correct alternative:
`int ((x^3 + 3x^2 + 3x + 1))/(x + 1)^5 "d"x` =
Evaluate `int (2"e"^x + 5)/(2"e"^x + 1) "d"x`
Evaluate: `int (dx)/(2 + cos x - sin x)`
Evaluate: `int_-2^1 sqrt(5 - 4x - x^2)dx`
`int 1/(x^2 + 1)^2 dx` = ______.
Evaluate: `int (2x^2 - 3)/((x^2 - 5)(x^2 + 4))dx`
Find: `int x^4/((x - 1)(x^2 + 1))dx`.
Evaluate.
`int (5x^2 - 6x + 3) / (2x -3) dx`
If \[\int\frac{2x+3}{(x-1)(x^{2}+1)}\mathrm{d}x\] = \[=\log_{e}\left\{(x-1)^{\frac{5}{2}}\left(x^{2}+1\right)^{a}\right\}-\frac{1}{2}\tan^{-1}x+\mathrm{A}\] where A is an arbitrary constant, then the value of a is
