Advertisements
Advertisements
प्रश्न
Evaluate:
`inte^x "cosec" x(1 - cot x)dx`
उत्तर
`inte^x "cosec" x(1 - cot x)dx`
= `int e^x("cosec" x - "cosec" x . cot x)dx`
= `int e^x("cosec" x + (-"cosec" x . cot x))dx` ...`[∵ inte^x(f(x) + f^'(x))dx = e^xf(x) + c]`
= `e^x"cosec" x + c`
APPEARS IN
संबंधित प्रश्न
Evaluate `int_0^(pi)e^2x.sin(pi/4+x)dx`
Integrate the function in (sin-1x)2.
Integrate the function in x sec2 x.
Integrate the function in `sin^(-1) ((2x)/(1+x^2))`.
Evaluate the following : `int x^2.log x.dx`
Evaluate the following : `int x.cos^3x.dx`
Integrate the following functions w.r.t. x : `sqrt(2x^2 + 3x + 4)`
Integrate the following functions w.r.t. x : `log(1 + x)^((1 + x)`
Integrate the following with respect to the respective variable : `(3 - 2sinx)/(cos^2x)`
Integrate the following with respect to the respective variable : cos 3x cos 2x cos x
Integrate the following w.r.t.x : `log (1 + cosx) - xtan(x/2)`
Integrate the following w.r.t.x : `(1)/(x^3 sqrt(x^2 - 1)`
Integrate the following w.r.t.x : log (x2 + 1)
`int (cos2x)/(sin^2x cos^2x) "d"x`
`int (x^2 + x - 6)/((x - 2)(x - 1)) "d"x` = x + ______ + c
`int logx/(1 + logx)^2 "d"x`
`int "e"^x [x (log x)^2 + 2 log x] "dx"` = ______.
If u and v ore differentiable functions of x. then prove that:
`int uv dx = u intv dx - int [(du)/(d) intv dx]dx`
Hence evaluate `intlog x dx`
Evaluate: `int_0^(pi/4) (dx)/(1 + tanx)`
`int e^x [(2 + sin 2x)/(1 + cos 2x)]dx` = ______.
Find: `int e^(x^2) (x^5 + 2x^3)dx`.
The integrating factor of `ylogy.dx/dy+x-logy=0` is ______.
`int(f'(x))/sqrt(f(x)) dx = 2sqrt(f(x))+c`
Evaluate:
`int((1 + sinx)/(1 + cosx))e^x dx`
Evaluate:
`inte^x sinx dx`
Evaluate:
`int1/(x^2 + 25)dx`
If ∫(cot x – cosec2 x)ex dx = ex f(x) + c then f(x) will be ______.
