Advertisements
Advertisements
प्रश्न
Prove the following:
`int_0^(pi/4) 2 tan^3 xdx = 1 - log 2`
उत्तर
`int_0^(pi/4) 2 tan^3 x dx`
`= int_0^(pi/4) 2 tan x* tan^2 x dx`
`= int_0^(pi/4) 2 tan x (sec^2 x - 1) dx`
`= 2 int_0^(pi/4) (tan x) sec^2 x dx - 2 int_0^(pi/2) tan x dx`
`= 2 [(tan^2 x)/2]_0^(pi/4) - 2 [- log |cos x|] _0^(pi/4)`
`= (tan^2 pi/4 - tan^2 0) + 2 (log cos pi/4 - log cos 0)`
`= (1 - 0) + 2 (log 1/ sqrt2 - log 1)`
`= 1 + 2 (log 1 - log sqrt 2 - log 1)`
`= 1 + 2 xx (-1/2 log 2)`
= 1 - log 2
APPEARS IN
संबंधित प्रश्न
Evaluate `int_1^3(e^(2-3x)+x^2+1)dx` as a limit of sum.
Evaluate the definite integral:
`int_(pi/2)^pi e^x ((1-sinx)/(1-cos x)) dx`
Evaluate the definite integral:
`int_(pi/6)^(pi/3) (sin x + cosx)/sqrt(sin 2x) dx`
Evaluate the definite integral:
`int_0^1 dx/(sqrt(1+x) - sqrtx)`
Evaluate the definite integral:
`int_1^4 [|x - 1|+ |x - 2| + |x -3|]dx`
Prove the following:
`int_(-1)^1 x^17 cos^4 xdx = 0`
Prove the following:
`int_0^(pi/2) sin^3 xdx = 2/3`
Prove the following:
`int_0^1sin^(-1) xdx = pi/2 - 1`
\[\int\frac{1}{x} \left( \log x \right)^2 dx\]
\[\int\limits_0^1 \left( x e^x + \cos\frac{\pi x}{4} \right) dx\]
Evaluate the following integral:
Evaluate the following integrals as limit of sums:
\[\int\frac{\sqrt{\tan x}}{\sin x \cos x} dx\]
Using L’Hospital Rule, evaluate: `lim_(x->0) (8^x - 4^x)/(4x
)`
If f and g are continuous functions in [0, 1] satisfying f(x) = f(a – x) and g(x) + g(a – x) = a, then `int_0^"a" "f"(x) * "g"(x)"d"x` is equal to ______.
Evaluate the following as limit of sum:
`int _0^2 (x^2 + 3) "d"x`
Evaluate the following:
`int_0^2 ("d"x)/("e"^x + "e"^-x)`
Evaluate the following:
`int_0^(pi/2) (tan x)/(1 + "m"^2 tan^2x) "d"x`
Evaluate the following:
`int_0^(1/2) ("d"x)/((1 + x^2)sqrt(1 - x^2))` (Hint: Let x = sin θ)
The value of `int_(-pi)^pi sin^3x cos^2x "d"x` is ______.
The value of `lim_(x -> 0) [(d/(dx) int_0^(x^2) sec^2 xdx),(d/(dx) (x sin x))]` is equal to
Left `f(x) = {{:(1",", "if x is rational number"),(0",", "if x is irrational number"):}`. The value `fof (sqrt(3))` is
