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
Evaluate: `∫8/((x+2)(x^2+4))dx`
Integrate the rational function:
`x/((x + 1)(x+ 2))`
Integrate the rational function:
`(1 - x^2)/(x(1-2x))`
Integrate the rational function:
`2/((1-x)(1+x^2))`
Integrate the rational function:
`(3x -1)/(x + 2)^2`
Integrate the rational function:
`1/(x(x^4 - 1))`
Evaluate : `∫(x+1)/((x+2)(x+3))dx`
Find `int (2cos x)/((1-sinx)(1+sin^2 x)) dx`
Integrate the following w.r.t. x : `(x^2 + 2)/((x - 1)(x + 2)(x + 3)`
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 : `(1)/(x^3 - 1)`
Integrate the following w.r.t. x : `((3sin - 2)*cosx)/(5 - 4sin x - cos^2x)`
Integrate the following w.r.t. x : `(1)/(2sinx + sin2x)`
Integrate the following w.r.t. x: `(x^2 + 3)/((x^2 - 1)(x^2 - 2)`
Integrate the following with respect to the respective variable : `cot^-1 ((1 + sinx)/cosx)`
Integrate the following w.r.t.x : `x^2/sqrt(1 - x^6)`
Integrate the following w.r.t.x : `sqrt(tanx)/(sinx*cosx)`
Evaluate:
`int (2x + 1)/(x(x - 1)(x - 4)) dx`.
Evaluate:
`int x/((x - 1)^2(x + 2)) dx`
Evaluate: `int "3x - 2"/(("x + 1")^2("x + 3"))` dx
Evaluate: `int 1/("x"("x"^"n" + 1))` dx
State whether the following statement is True or False.
If `int (("x - 1") "dx")/(("x + 1")("x - 2"))` = A log |x + 1| + B log |x - 2| + c, then A + B = 1.
`int "e"^(3logx) (x^4 + 1)^(-1) "d"x`
`int x^2sqrt("a"^2 - x^6) "d"x`
`int 1/(x(x^3 - 1)) "d"x`
`int (7 + 4x + 5x^2)/(2x + 3)^(3/2) dx`
`int 1/(2 + cosx - sinx) "d"x`
`int "e"^(sin^(-1_x))[(x + sqrt(1 - x^2))/sqrt(1 - x^2)] "d"x`
`int "e"^x ((1 + x^2))/(1 + x)^2 "d"x`
`int (x + sinx)/(1 - cosx) "d"x`
`int ((2logx + 3))/(x(3logx + 2)[(logx)^2 + 1]) "d"x`
`int (5(x^6 + 1))/(x^2 + 1) "d"x` = x5 – ______ x3 + 5x + c
Evaluate `int x log x "d"x`
`int 1/(4x^2 - 20x + 17) "d"x`
`int (3"e"^(2"t") + 5)/(4"e"^(2"t") - 5) "dt"`
Evaluate the following:
`int x^2/(1 - x^4) "d"x` put x2 = t
Evaluate the following:
`int (x^2"d"x)/(x^4 - x^2 - 12)`
Evaluate the following:
`int (x^2 "d"x)/((x^2 + "a"^2)(x^2 + "b"^2))`
Evaluate the following:
`int_"0"^pi (x"d"x)/(1 + sin x)`
Let g : (0, ∞) `rightarrow` R be a differentiable function such that `int((x(cosx - sinx))/(e^x + 1) + (g(x)(e^x + 1 - xe^x))/(e^x + 1)^2)dx = (xg(x))/(e^x + 1) + c`, for all x > 0, where c is an arbitrary constant. Then ______.
`int 1/(x^2 + 1)^2 dx` = ______.
Evaluate`int(5x^2-6x+3)/(2x-3)dx`
Evaluate:
`int 2/((1 - x)(1 + x^2))dx`
Evaluate:
`int (x + 7)/(x^2 + 4x + 7)dx`
