Advertisements
Advertisements
प्रश्न
If `int_0^a1/(4+x^2)dx=pi/8` , find the value of a.
Advertisements
उत्तर
given that `int_0^a1/(4+x^2)dx=pi/8`
We need to find the value of a.
`Let I=int_0^a1/(4+x^2)dx=pi/8`
`Thus,I=1/2(tan^(-1)(x/2))_0^a=pi/8`
`=>1/2 tan^(-1)(a/2)=pi/8`
`=>tan^(-1)(a/2)=pi/4`
`=>a/2=tan(pi/4)`
`=>a/2=1`
`a=2`
APPEARS IN
संबंधित प्रश्न
Evaluate: `int1/(xlogxlog(logx))dx`
Evaluate : `int1/(3+5cosx)dx`
Evaluate `∫_0^(3/2)|x cosπx|dx`
Evaluate the integral by using substitution.
`int_0^1 sin^(-1) ((2x)/(1+ x^2)) dx`
Evaluate the integral by using substitution.
`int_0^2 dx/(x + 4 - x^2)`
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate the following integral:
Evaluate
\[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\]
Evaluate the following integral:
Evaluate :
Evaluate: `int_1^5{|"x"-1|+|"x"-2|+|"x"-3|}d"x"`.
Find: `int_ (3"x"+ 5)sqrt(5 + 4"x"-2"x"^2)d"x"`.
`int_0^3 1/sqrt(3x - x^2)"d"x` = ______.
`int_0^(pi4) sec^4x "d"x` = ______.
Find: `int (dx)/sqrt(3 - 2x - x^2)`
The value of `int_0^1 (x^4(1 - x)^4)/(1 + x^2) dx` is
Evaluate: `int_0^(π/2) sin 2x tan^-1 (sin x) dx`.
Evaluate:
`int (1 + cosx)/(sin^2x)dx`
