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
find : `int(3x+1)sqrt(4-3x-2x^2)dx`
Find:
`int(x^3-1)/(x^3+x)dx`
Evaluate:
`int((x+3)e^x)/((x+5)^3)dx`
Integrate the function `(3x^2)/(x^6 + 1)`
Integrate the function `1/sqrt((2-x)^2 + 1)`
Integrate the function `1/sqrt(9 - 25x^2)`
Integrate the function `x^2/sqrt(x^6 + a^6)`
Integrate the function `1/sqrt(x^2 +2x + 2)`
Integrate the function `1/sqrt(8+3x - x^2)`
Integrate the function `(4x+ 1)/sqrt(2x^2 + x - 3)`
Integrate the function `(5x - 2)/(1 + 2x + 3x^2)`
Integrate the function `(6x + 7)/sqrt((x - 5)(x - 4))`
`int dx/(x^2 + 2x + 2)` equals:
Integrate the function:
`sqrt(x^2 + 4x - 5)`
Integrate the function:
`sqrt(1+ 3x - x^2)`
Evaluate : `int_2^3 3^x dx`
Find `int (2x)/(x^2 + 1)(x^2 + 2)^2 dx`
\[\int\frac{8x + 13}{\sqrt{4x + 7}} \text{ dx }\]
Find : \[\int\left( 2x + 5 \right)\sqrt{10 - 4x - 3 x^2}dx\] .
`int (a^x - b^x)^2/(a^xb^x)dx` equals ______.
Evaluate \[\int \frac{dx}{\sqrt{x^2-a^2}}\].
Which method transforms \[\int\frac{dx}{ax^2+bx+c}\] or \[\int\frac{dx}{\sqrt{ax^2+bx+c}}\] into an expression compatible with standard formulae?
For \[\int\frac{px+q}{ax^2+bx+c}\,dx\], how is \[px+q\] written in relation to the derivative of the denominator?
Evaluate \[\int\frac{dx}{x^2-6x+13}\].
Which statement is correct when integrating an expression that is not immediately a standard integral?
