Advertisements
Advertisements
प्रश्न
Evaluate : `int (sec^2 x)/(tan^2 x + 4)` dx
Advertisements
उत्तर
Let I = `int (sec^2 x)/(tan^2 x + 4)` dx
Put tan x = t
`sec^2 x dx = dt`
I = `int dt/[ t^2 + 2^2 ]`
I = `1/2 tan^-1 (t/2) + c`
`( ∴ int 1/[ x^2 + a^2] dx = 1/a tan^-1 x/a + c)`
I = `1/2 tan^-1(tan x/2) + c`
APPEARS IN
संबंधित प्रश्न
If x = f(t), y = g(t) are differentiable functions of parammeter ‘ t ’ then prove that y is a differentiable function of 'x' and hence, find dy/dx if x=a cost, y=a sint
If x=at2, y= 2at , then find dy/dx.
If x=a sin 2t(1+cos 2t) and y=b cos 2t(1−cos 2t), find `dy/dx `
If x = a sin 2t (1 + cos2t) and y = b cos 2t (1 – cos 2t), find the values of `dy/dx `at t = `pi/4`
Find the value of `dy/dx " at " theta =pi/4 if x=ae^theta (sintheta-costheta) and y=ae^theta(sintheta+cos theta)`
If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.
x = a cos θ, y = b cos θ
If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.
x = 4t, y = `4/y`
If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.
x = a (θ – sin θ), y = a (1 + cos θ)
If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.
x = `a(cos t + log tan t/2)`, y = a sin t
If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.
x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ)
If x = `sqrt(a^(sin^(-1)t))`, y = `sqrt(a^(cos^(-1)t))` show that `dy/dx = - y/x`.
If `x = acos^3t`, `y = asin^3 t`,
Show that `(dy)/(dx) =- (y/x)^(1/3)`
x = `"t" + 1/"t"`, y = `"t" - 1/"t"`
x = `"e"^theta (theta + 1/theta)`, y= `"e"^-theta (theta - 1/theta)`
Differentiate `tan^-1 ((sqrt(1 + x^2) - 1)/x)` w.r.t. tan–1x, when x ≠ 0
If x = t2, y = t3, then `("d"^2"y")/("dx"^2)` is ______.
Derivative of x2 w.r.t. x3 is ______.
