Question

The value of int\ sqrt(4-tan^2 x) \ sec^2 x \ dx equals

Answer 1

EXPLANATION.

$\sf \implies \displaystyle \int \sqrt{(4 - tan^{2} x)} sec^{2}x \ dx.$

As we know that,

By applying substitution method in this question,

We can write equation as,

Let we assume that,

⇒ tan x = t.

Differentiate w.r.t x, we get.

⇒ sec²xdx = dt.

Put the values in the equation, we get.

$\sf \implies \displaystyle \int \sqrt{4 - t^{2} } \ dt.$

$\sf \implies \displaystyle \int \sqrt{(2)^{2} - t^{2} } \ dt.$

As we know that,

Formula of :

$\sf \implies \displaystyle \int \sqrt{a^{2} - x^{2} } \ dx = \dfrac{x}{2} \sqrt{a^{2} - x^{2} } + \dfrac{a^{2} }{2} sin^{-1} \bigg( \dfrac{x}{a} \bigg) + C.$

Using this formula in the equation, we get.

$\sf \implies \displaystyle \dfrac{t}{2} \sqrt{4 - t^{2} } + \dfrac{(2)^{2} }{2} sin^{-1} \bigg( \dfrac{t}{2} \bigg) + C.$

Put the value of t = tan x in the equation, we get.

$\sf \implies \displaystyle \dfrac{tan(x)}{2} \sqrt{4 - tan^{2}x } + 2 \ sin^{-1} \bigg( \dfrac{tan(x)}{2} \bigg) + C.$

                                                                                                                   

MORE INFORMATION.

Standard integrals.

(1) = ∫0.dx = c.

(2) = ∫1.dx = x + c.

(3) = ∫k dx = kx + c, (k ∈ R).

(4) = ∫xⁿdx = xⁿ⁺¹/n + 1 + c, (n ≠ - 1).

(5) = ∫dx/x = ㏒(x) + c.

(6) = ∫eˣdx = eˣ + c.

(7) = ∫aˣdx = aˣ/㏒(a) + c = aˣ ㏒(e) + c.