Question

The value of ∫[(1-cot^n-2 x)dx/(tan x + cot x. cot^n-2 x)] is equal to​

Answer 1

Step-by-step explanation:

$\bf \underline{Solution-}$

$\sf \implies \displaystyle \int \dfrac{(1 - cot^{n - 2}x)dx }{tan x + cot x. cot^{n - 2} x}$

As we know that,

We can write equation as,

$\sf \implies \displaystyle \int \dfrac{1 - \dfrac{cos^{n - 2} x}{sin^{n - 2}x } }{\dfrac{sin x}{cos x} + \dfrac{cos x}{sin x}. \dfrac{cos^{n - 2}x }{sin^{n - 2} x} } dx.$

$\sf \implies \displaystyle \int \dfrac{\bigg( \dfrac{sin^{n - 2}x - cos^{n - 2} x }{sin^{n - 2} x}\bigg) dx }{\bigg(\dfrac{sin x}{cos x}+ \dfrac{cos^{n - 1} x}{sin^{n - 1} x}\bigg) }$

$\sf \implies \displaystyle \int \dfrac{\bigg( \dfrac{sin^{n - 2}x - cos^{n - 2}x }{sin^{n - 2} x} \bigg) dx}{\dfrac{(sin x)(sin^{n - 1}x)+ (cos x)(cos^{n - 1} x)}{(cos x)(sin^{n - 1} x)} }$

$\sf \implies \displaystyle \int \bigg[ \dfrac{sin^{n - 2}x - cos^{n - 2} x}{sin^{n - 2} x} \times \dfrac{(cos x)(sin^{n - 1} x)}{(sin^{n} x + cos^{n} x)} \bigg] dx$

$\sf \implies \displaystyle \int \bigg[ \dfrac{sin^{n - 2} x - cos^{n - 2}x }{\bigg(\dfrac{sin^{n - 1}x }{sin x} \bigg)} \times \dfrac{(cos x)(sin^{n - 1} x)}{(sin^{n}x + cos^{n} x) } \bigg] dx$

$\sf \implies \displaystyle \int \bigg[ \dfrac{sin^{n - 2}x - cos^{n - 2} x }{sin^{n - 1} x} \times \dfrac{(cos x)(sin x)(sin^{n - 1}x) }{(sin^{n} x + cos^{n} x)} \bigg] dx$

$\sf \implies \displaystyle \int \bigg[ \dfrac{(sin^{n - 2}x - cos^{n - 2}x) \times (cos x)(sin x) }{(sin^{n} x + cos^{n} x)} \bigg] dx$

Now, Let we assume that,

⇒ sinⁿx + cosⁿx = t.

Differentiate w.r.t x, we get.

⇒ n⁻¹(sinⁿ⁻²x - cosⁿ⁻²x)(sin x cos x) = dt.

Put the values in the equation, we get.

$\sf \implies \displaystyle \frac{1}{n} \int \dfrac{1}{t} dt.$

Put the values of t = sinⁿx + cosⁿx in the equation, we get.

$\sf \implies \displaystyle \frac{1}{n} log |sin^{n} x + cos^{n} x | + C.$

$\sf \implies \displaystyle \int \dfrac{(1 - cot^{n - 2}x)dx }{tan x + cot x. cot^{n - 2} x} = \dfrac{1}{n} log|sin^{n} x + cos^{n} x| + C.$