Question

Let In = ∫ tan^n x dx (n > 1). If I4 + I6 = a tan^5 x + bx^5 + C,

Answer 1

EXPLANATION.

⇒ Iₙ = ∫tanⁿdx (n > 1).

⇒ I₄ + I₆ = a. tan⁵x + bx⁵ + c.

As we know that,

Put the value of n = 4 and n = 6 in the equation, we get.

⇒ I₄ + I₆ = ∫[tan⁴ x + tan⁶ x]dx.

⇒ I₄ + I₆ = ∫[tan⁴ x (1 + tan² x)]dx.

⇒ I₄ + I₆ = ∫[tan⁴ x . sec² x]dx.

By using substitution method in this equation, we get.

Let, we assume that,

⇒ tan x = t.

Differentiate both sides w.r.t x, we get.

⇒ sec²x dx = dt.

Put the value in the equation, we get.

⇒ I₄ + I₆ = ∫t⁴ dt.

⇒ I₄ + I₆ = t⁵/5 + c.

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

⇒ I₄ + I₆ = tan⁵ x/5 + c.

As we can see that,

Value of a = 1/5  and  b = 0.

                                                                                                                       

MORE INFORMATION.

(1) = ∫sin x dx = - cos x + c.

(2) = ∫cos x dx = sin x + c.

(3) = ∫tan x dx = ㏒(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

★ QUESTION ★

Let In = ∫ tan^n x dx (n > 1). If I4 + I6 = a tan^5 x + bx^5 + C,

✵ SOLUTION ✵

GIVEN IN ATTACHMENT .

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