Advertisements
Advertisements
प्रश्न
Find:
`int_(-pi/4)^0 (1+tan"x")/(1-tan"x") "dx"`
Advertisements
उत्तर
Let
` "I" = int_(-pi/4)^0 (1+tan"x")/(1-tan"x") "dx"`
` "I" = int_(-pi/4)^0 (1+sin"x"/cos"x")/(1-sin"x"/cos"x")"dx"`
` "I" = int_(-pi/4)^0 (cos"x"+sin"x")/(cos"x"-sin"x")"dx"`
Let
cos x - sin x = t
- (sin x + cos x) dx = dt
For x `=(-pi)/4,"t" = sqrt2` and x = 0, t = 1
`"I" = -int_(sqrt2)^1 "dt"/"t"`
`"I" = int_1^sqrt2 "dt"/"t"`
` = ["lnt"]_1^sqrt2`
` = "ln"sqrt2`
APPEARS IN
संबंधित प्रश्न
Evaluate : ` int x^2/((x^2+4)(x^2+9))dx`
Find:
`int(x^3-1)/(x^3+x)dx`
Integrate the function `1/sqrt((2-x)^2 + 1)`
Integrate the function `1/sqrt(9 - 25x^2)`
Integrate the function `(3x)/(1+ 2x^4)`
Integrate the function `(x - 1)/sqrt(x^2 - 1)`
Integrate the function `1/sqrt(x^2 +2x + 2)`
Integrate the function `1/sqrt((x -1)(x - 2))`
Integrate the function `1/sqrt((x - a)(x - b))`
Integrate the function `(4x+ 1)/sqrt(2x^2 + x - 3)`
`int dx/(x^2 + 2x + 2)` equals:
Integrate the function:
`sqrt(x^2 + 4x +1)`
Integrate the function:
`sqrt(1+ 3x - x^2)`
Integrate the function:
`sqrt(1+ x^2/9)`
`int sqrt(1+ x^2) dx` is equal to ______.
`int sqrt(x^2 - 8x + 7) dx` is equal to ______.
Evaluate : `int_2^3 3^x dx`
Find `int (2x)/(x^2 + 1)(x^2 + 2)^2 dx`
Integration of \[\frac{1}{1 + \left( \log_e x \right)^2}\] with respect to loge x is
\[\int\frac{8x + 13}{\sqrt{4x + 7}} \text{ dx }\]
Find : \[\int\left( 2x + 5 \right)\sqrt{10 - 4x - 3 x^2}dx\] .
Find: `int (dx)/(x^2 - 6x + 13)`
