Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int sqrt(tanx) "d"x` (Hint: Put tanx = t2)
Advertisements
उत्तर
Let I = `int sqrt(tanx) "d"x`
Put tan x = t2
⇒ sec2x dx = 2t dt
∴ I = `int "t" * (2"t")/(sec^2x) "dt"`
= `2 int "t"^2/(1 + "t"^4) "dt"`
= `int (("t"^2 + 1) + ("t"^2 - 1))/((1 + "t"^4)) "dt"`
= `int ("t"^2 + 1)/(1 + "t"^4) "dt" + int ("t"^2 - 1)/(1 + "t"^4) "dt"`
= `int (1 + 1/"t"^2)/("t"^2 + 1/"t"^2) "dt" + int (1 - 1/"t"^2)/("t"^2 + 1/"t"^2) "dt"`
= `int (1 + 1/"t"^2)/(("t" - 1/"t")^2 + 2)"dt" + int (1 - 1/"t"^2)/(("t" + 1/"t")^2 - 2)"dt"`
Put u = `"t" - 1/"t"`
⇒ du = `(1 + 1/"t"^2)"dt"` in first integral
And put v = `"t" + 1/"t"`
⇒ dv = `(1 - 1/"t"^2)"dt"` in second integral
∴ I = `int "du"/("u"^2 + (sqrt(2)^2)) + int "dv"/("v"^2 - (sqrt(2)^2))`
= `1/sqrt(2) tan^-1 "u"/sqrt(2) + 1/(2sqrt(2)) log|("v" - sqrt(2))/("v" + sqrt(2))| + "C"`
= `1/sqrt(2) tan^-1 ("t" - 1/"t")/sqrt(2) + 1/(2sqrt(2)) log |("t" + 1/"t" - sqrt(2))/("t" + 1/"t" + sqrt(2))| + "C"`
= `1/sqrt(2) tan^-1 ("t"^2 - 1)/(sqrt(2)"t") + 1/(2sqrt(2)) log |("t"^2 + 1 - sqrt(2)"t")/("t"^2 + 1 + sqrt(2)"t")| + "C"`
= `1/sqrt(2) tan^-1 ((tanx - 1)/sqrt(2tan x)) + 1/(2sqrt(2)) log |(tan x - sqrt(2 tanx) + 1)/(tan x + sqrt(2 tan x) + 1)| + "C"`
APPEARS IN
संबंधित प्रश्न
Find : `int x^2/(x^4+x^2-2) dx`
Find: `I=intdx/(sinx+sin2x)`
Evaluate: `∫8/((x+2)(x^2+4))dx`
Integrate the rational function:
`x/((x^2+1)(x - 1))`
Integrate the rational function:
`x/((x -1)^2 (x+ 2))`
Integrate the rational function:
`1/(x^4 - 1)`
`int (dx)/(x(x^2 + 1))` equals:
Find `int(e^x dx)/((e^x - 1)^2 (e^x + 2))`
Find `int (2cos x)/((1-sinx)(1+sin^2 x)) dx`
Integrate the following w.r.t. x : `(1)/(x(x^5 + 1)`
Integrate the following w.r.t. x : `2^x/(4^x - 3 * 2^x - 4`
Integrate the following w.r.t. x : `(1)/(x^3 - 1)`
Integrate the following w.r.t. x : `(1)/(2sinx + sin2x)`
Integrate the following w.r.t. x : `(1)/(sin2x + cosx)`
Integrate the following w.r.t. x : `(5*e^x)/((e^x + 1)(e^(2x) + 9)`
Integrate the following with respect to the respective variable : `(6x + 5)^(3/2)`
Integrate the following w.r.t.x : `x^2/sqrt(1 - x^6)`
Integrate the following w.r.t.x : `(1)/(sinx + sin2x)`
Integrate the following w.r.t.x : `sec^2x sqrt(7 + 2 tan x - tan^2 x)`
Integrate the following w.r.t.x : `sqrt(tanx)/(sinx*cosx)`
Evaluate: `int 1/("x"("x"^"n" + 1))` dx
`int "dx"/(("x" - 8)("x" + 7))`=
`int x^7/(1 + x^4)^2 "d"x`
`int 1/(4x^2 - 20x + 17) "d"x`
`int (6x^3 + 5x^2 - 7)/(3x^2 - 2x - 1) "d"x`
Evaluate `int x^2"e"^(4x) "d"x`
Evaluate: `int (dx)/(2 + cos x - sin x)`
Evaluate: `int (2x^2 - 3)/((x^2 - 5)(x^2 + 4))dx`
Evaluate`int(5x^2-6x+3)/(2x-3)dx`
Evaluate.
`int (5x^2 - 6x + 3)/(2x - 3)dx`
Evaluate:
`int(2x^3 - 1)/(x^4 + x)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
When is a rational function called improper?
Which decomposition corresponds to \[\frac{\mathrm{p}x^{2}+\mathrm{q}x+\mathrm{r}}{(x-\mathrm{a})(x^{2}+\mathrm{b}x+\mathrm{c})}\]?
What is done after long division, before writing the appropriate partial-fraction decomposition?
For \[\frac{5x-5}{(x-2)(x-3)}=\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{x-3}\], which equation results after clearing denominators?
Evaluate \[\int\frac{x^{2}+1}{x^{2}-5x+6}\,dx\].
What numerator should be used for each distinct linear factor in a partial-fraction decomposition?
