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$\bf {Let \: I_n = \int { tan}^{n} \: x \: dx(n > 1)} \\ \\ \bf If \: I_4 + I_6 = a. {tan}^{5}x + b {x}^{5} + C$
Then Find Values of a & b
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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

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♣ Given :-

• $\bf { I_n = \int { tan}^{n} \: x \: dx(n > 1)}$

• $\bf If \: I_4 + I_6 = a. {tan}^{5}x + b {x}^{5} + C$

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♣ To Find :-

• Values of a and b

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♣ Answer :-

• $\sf a = \dfrac{1}{5}$

• b = 0

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♣ Step by step Explanation :-

We Have

${I_n = \int { \tan}^{n} \: \mathtt x \: d\mathtt x }$

$\therefore I_n + I_{n + 2} = \int { \tan}^{n} \: \mathtt x \: d\mathtt x + \int { \tan}^{n + 2} \: \mathtt x \: d\mathtt x \\\\ \sf= \int ({ \tan}^{n} \: \mathtt x + { \tan}^{n + 2} \: \mathtt x) \: d\mathtt x \\ \\ = \sf\int { \tan}^{n} \mathtt x \: (1 + { \tan}^{2} \mathtt x ) \: d\mathtt x \\ \\ = \sf \int { \tan}^{n} \mathtt x . { \sec}^{2} \mathtt x \: d\mathtt x \\ \\ :\longmapsto I_n + I_{n + 2} = \dfrac{{ \tan}^{n + 1} \mathtt x }{n + 1} +C$

Putting n = 4 We Get ,

$\pmb{I_4+I_6 = \dfrac{ { \tan}^{5}x}{5} + C }$

Comparing it with the Given Equation :

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf a = \frac{1}{5} } }}} \: \: \bold{and} \: \: \purple{ \large \underline {\boxed{{\bf b = 0} }}}$

$\LARGE\red{\mathfrak{  \text{W}hich \:\:is\:\: the\:\: required} }\\ \Huge \red{\mathfrak{ \text{ A}nswer.}}$

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♣ Related Calculations :-

$\sf \int { \tan}^{n} \mathtt x. { \sec}^{2} \mathtt x \: d\mathtt x \\ \\ \sf Putting \: \tan\mathtt x = t \\ \\ :\longmapsto \sf { \sec}^{2} \mathtt x \: d\mathtt x = dt \\ \\ :\longmapsto \sf \int { \tan}^{n} \mathtt x. { \sec}^{2} \mathtt x \: d\mathtt x = \int {t}^{n} \: dt \\ \\ = \sf\frac{ {t}^{n + 1} }{n + 1} \\ \\ = \sf\frac{ { \tan}^{n + 1} \mathtt x}{n + 1}$

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♣ Additional Information :-

$\large \bigstar \:   \underline{ \pmb{ \mathfrak{ Commonl \text y \: \: Used \: \: \text Integrals }}}:$

$\maltese \: \: \: \displaystyle \bf\int \dfrac{1}{x} \: dx =log |x + | C \\ \\ \maltese \: \: \: \displaystyle \bf\int {a}^{x} \: dx = \frac{ {a}^{x} }{log \: a} + C \\ \\ \maltese \: \: \: \displaystyle \bf\int sinx \: dx = - cosx + C \\ \\ \maltese \: \: \: \displaystyle \bf\int cosx \: dx =sinx + C \\ \\ \maltese \: \: \: \displaystyle \bf\int sec {}^{2} x \: dx =tanc + C \\ \\ \maltese \: \: \: \displaystyle \bf\int {cosec}^{2} x \: dx = - cotx + C \\ \\ \maltese \: \: \: \displaystyle \bf\int secx.tanx \: dx =secx + C \\ \\ \maltese \: \: \: \displaystyle \bf\int cosecx.cotx \: dx = - cosecx + C$

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