Question

do this please
$\huge \red \int \ \cot^{3} (x) .dx$
$\huge \bold \green{integration}$
​

Answer 1

EXPLANATION.

⇒ ∫cot³x dx.

As we know that,

We can write equation as,

⇒ ∫[cot(x).cot²(x)]dx.

As we know that,

Formula of :

⇒ 1 + cot²x = cosec²x.

⇒ cot²x = cosec²x - 1.

Using this formula in equation, we get.

⇒ ∫(cosec²x - 1).cot(x)dx.

⇒ ∫(cosec²x.cot(x)dx - ∫(cot(x)dx.

⇒ ∫[cos(x)/sin³(x)dx - ∫cot(x)dx.

From the first integration, we get.

⇒ ∫[cos(x)/sin³x]dx.

By using substitution method, we get.

Let we assume that,

⇒ sin(x) = t.

Differentiate w.r.t x, we get.

⇒ cos(x)dx = dt.

Put the values in the equation, we get.

⇒ ∫dt/t³ = -1/2t².

Put the value of t = sin(x) in equation, we get.

⇒ -1/2sin²x.

⇒ ∫[cos(x)/sin³(x)dx - ∫cot(x)dx.

⇒ -1/2sin²x - ㏒(sin x) + C.

                                                                                                                       

MORE INFORMATION.

Standard integrals.

(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) = ∫sec²x dx = tan x + c.

(10) = ∫cosec²x dx = - cot x + c.

Answer 2

Answer :

${ \underline{ \boxed{ \bf{ \pink{ \: - \frac{1}{2 {sin}^{2}x } - ln(sin \: x) +C \: }}}}}$

Explanation :

$\dashrightarrow \rm \: \int \: {cot}^{3} (x) \: dx$

$\dashrightarrow \rm \: \int \:cot \: x \: {cot}^{2} \: x \: \: dx$

$\bf\underline{ \star\:\:{Using \: \: the \: \: Pythagorean \: \: Identities : }}$

$\boxed{ {cot}^{2} \: x = {csc}^{2} \: x - 1}$

$\dashrightarrow \rm \: \int( {csc}^{2} x - 1)cot \: x \: dx$

★Expand the above Equation :

$\dashrightarrow \rm \: \int \: cot \: x \: {csc}^{2} \: dx \: - \int \: cot \: x \: dx$

★ Using the sum Rule :

$\boxed{ \sf {\int \: f(x) + g(x)dx = \int \: f(x)dx \: + \int \: g(x)dx}}$

$\dashrightarrow \rm \: \int \: cot \: x \: {csc}^{2} \: x \: dx \: - \int \: cot \: x \: dx$

$\dashrightarrow \rm \: \int \frac{cos \: x}{ {sin}^{3} x} \: dx \: - \int \: cot \: x \: dx$

★ Using integration by substitution on

$\int \: \frac{cos \: x}{ {sin}^{3} x} \: dx$

Let u = sin x , du = cos x dx .

★Using u and du in above step !!

$\dashrightarrow \rm \: \int \frac{1}{ {u}^{3} } du$

★By using Power Rule :

$\int \: {x}^{n}dx \: = \frac{ {x}^{n + 1} }{ {x}^{n + 1} } + C$

$\dashrightarrow \rm \: - \frac{1}{ {2u}^{2} }$

★ Substitute u = sin x

$\dashrightarrow \rm \: \frac{1}{ {2 \: sin \: }^{2} x}$

★ Now rewrite the integral with completed substitution.

$\dashrightarrow \rm \: \frac{1}{ {2 \: sin \: }^{2} x} \: - \int \: cot \: x \: dx$

★ Use Trigonometric integration .

cot x is In (sin x)

$\dashrightarrow \rm \: \frac{1}{ {2 \: sin}^{2}x } - ln(sin \: x)$

★ Add the constant :

$\dashrightarrow \rm \: \frac{1}{ {2 \: sin}^{2}x } - ln(sin \: x) +C$