Question

integral of sec^x/✓(tan^x+4) ​

Answer 1

EXPLANATION.

$\implies \displaystyle \int \dfrac{sec^{2}x dx}{\sqrt{tan^{2}x + 4 } }$

As we know that,

Apply substitution method in this equation, we get.

Let, we assume that.

⇒ tan x = t.

Differentiate w.r.t x, we get.

⇒ sec²x dx = dt.

Put the values in the equation, we get.

$\implies \displaystyle \int \dfrac{dt}{\sqrt{t^{2} + 4} }$

$\implies \displaystyle \int \dfrac{dt}{\sqrt{(t)^{2} + (2)^{2} } }$

As we know that,

Formula of :

$\implies \displaystyle \int \dfrac{dx}{\sqrt{x^{2} + a^{2} } } = log \bigg| x + \sqrt{x^{2} + a^{2} } \bigg| + C$

Using this formula in this equation, we get.

$\implies \displaystyle log \bigg| t + \sqrt{t^{2} + (2)^{2} } \bigg| + C$

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

$\implies \displaystyle log \bigg| tan (x) + \sqrt{tan^{2}x + 4 } \bigg| + C$

                                                                                                                         

MORE INFORMATION.

Standard integrals.

(1) = ∫sin x = - cos x + c.

(2) = ∫cos x = sin x + c.

(3) = ∫tan x = ㏒(sec x) + c = - ㏒(cos x) + c.

(4) = ∫cot x dx = ㏒(sin x) + c.

(5) = ∫sec x dx = ㏒(sec x + tan x) + c = - ㏒(sec x - tan x) + c = ㏒ tan(π/4 + x/2) + c.

(6) = ∫cosec x dx = - ㏒(cosec x + cot x) + c = ㏒(cosec x - cot x) + c = ㏒ tan(x/2) + c.

(7) = ∫sec x tan x dx = sec x + c.

(8) = ∫cosec x cot x dx = - cosec x + c.

(9) = ∫sec²xdx = tan x + c.

(10) = ∫cosec²xdx = - cot x + c.

Answer 2

ANSWER GIVEN IN ATTACHMENT.