Question

evaluate the following integration ​

Answer 1

EXPLANATION.

$\sf \implies \displaystyle\int\limits^\frac{\pi}{4} _\frac{-\pi}{4} {\dfrac{x + \dfrac{\pi}{4} }{2 - cos2x} } \, dx$

As we know that,

We can write equation as,

$\sf \implies \displaystyle\int\limits^\frac{\pi}{4} _\frac{-\pi}{4} \dfrac{x}{2 - cos2x} dx \ \ \ + \displaystyle\int\limits^\frac{\pi}{4} _\frac{-\pi}{4} \dfrac{\dfrac{\pi}{4} }{2 - cos2x} dx$

As we know that,

Formula of :

$\sf \implies \displaystyle\int_{-a}^{a} {f(x)} \, dx = 0, if \ f(-x) = -f(x) \ \ that \ \ is \ \ if \ \ \ f(x) \ \ is \ \ odd.$

$\sf \implies \displaystyle\int_{-a}^{a} {f(x)} \, dx = 2 \displaystyle\int\limits^a_0 {f(x)} \, dx if \ f(-x) = f(x) \ that \ is \ f(x) \ is \ even.$

Using this formula in equation, we get.

$\sf \implies 0 \ + 2 \displaystyle\int\limits^\frac{\pi}{4} _0 \dfrac{\dfrac{\pi}{4} }{2 - cos2x} dx.$

$\sf \implies 2 \times \dfrac{\pi}{4} \displaystyle\int\limits^\frac{\pi}{4} _0 \dfrac{dx}{2 - cos2x}$

$\sf \implies \dfrac{\pi}{2} \displaystyle\int\limits^\frac{\pi}{4} _0 \dfrac{dx}{2 - cos2x}$

As we know that,

Formula of :

⇒ cos2x = 1 - 2sin²x.

Put the formula in equation, we get.

$\sf \implies \dfrac{\pi}{2} \displaystyle\int\limits^\frac{\pi}{4} _0 \dfrac{dx}{2 - (1 - 2sin^{2}x) }$

$\sf \implies \dfrac{\pi}{2} \displaystyle\int\limits^\frac{\pi}{4} _0 \dfrac{dx}{1 + 2sin^{2} x}$

Divide and multiply this equation by cos²x, we get.

$\sf \implies \dfrac{\pi}{2} \displaystyle\int\limits^\frac{\pi}{4} _0 \dfrac{\dfrac{1}{cos^{2}x } }{\dfrac{1 + 2sin^{2}x }{cos^{2} x} } dx$

$\sf \implies \dfrac{\pi}{2} \displaystyle\int\limits^\frac{\pi}{4} _0 \dfrac{sec^{2}x }{\dfrac{1}{cos^{2}x } + \dfrac{2sin^{2} x}{cos^{2}x } } dx.$

$\sf \implies \dfrac{\pi}{2} \displaystyle\int\limits^\frac{\pi}{4} _0 \dfrac{sec^{2} x}{sec^{2}x + 2tan^{2} x } dx.$

As we know that,

Formula of :

⇒ sec²x = 1 + tan²x.

Put the value of sec²x in equation, we get.

$\sf \implies \dfrac{\pi}{2} \displaystyle\int\limits^\frac{\pi}{4} _0 \dfrac{sec^{2} x}{1 + tan^{2}x + 2tan^{2} x}dx.$

$\sf \implies \dfrac{\pi}{2} \displaystyle\int\limits^\frac{\pi}{4} _0 \dfrac{sec^{2} x}{1 + 3tan^{2}x }dx.$

Using the substitution method in this equation, we get.

Let we assume that,

⇒ tan x = t.

Differentiate w.r.t x, we get.

⇒ sec²x dx = dt.

As we know that,

in definite integration if we apply substitution method then limit will also change.

Put the value of x = 0, we get.

⇒ tan(0) = t.

⇒ t = 0.

Put the value of x = π/4  in equation, we get.

⇒ tan(π/4) = t.

⇒ tan45° = t.

⇒ t = 1.

Now the new limit is vary from 0 to 1.

$\sf \implies \dfrac{\pi}{2} \displaystyle\int\limits^1 _0 \dfrac{dt}{1 + 3t^{2} }$

$\sf \implies \dfrac{\pi}{2} \displaystyle\int\limits^1_0 \dfrac{dt}{3\bigg(\dfrac{1}{3}+ t^{2} \bigg) }.$

$\sf \implies \dfrac{\pi}{6} \displaystyle\int\limits^1_0 \dfrac{dt}{\bigg(\dfrac{1}{\sqrt{3} }\bigg)^{2} + t^{2} }$

As we know that,

Formula of :

⇒ ∫dx/x² + a² = 1/a tan⁻¹(x/a) + c.

Using this formula in equation, we get.

$\sf \implies \dfrac{\pi}{6} \bigg[\dfrac{1}{\dfrac{1}{\sqrt{3} } } \ tan^{-1} \bigg(\dfrac{t}{\dfrac{1}{\sqrt{3} } } \bigg) \bigg]_0^1$

$\sf \implies \dfrac{\pi}{6} \bigg[\sqrt{3} \ tan^{-1} \sqrt{3} t \bigg]_0^1$

$\sf \implies \dfrac{\sqrt{3} \pi}{6} \bigg[ \ tan^{-1} \sqrt{3} t \bigg]_0^1$

As we know that,

First we put upper limit then put the lower limit, we get.

$\sf \implies \dfrac{\sqrt{3} \pi}{6} \bigg[tan^{-1} \sqrt{3} (1) \ - \ tan^{-1} (0) \bigg]$

$\sf \implies \dfrac{\sqrt{3} \pi}{6} \bigg[\dfrac{\pi}{3} \bigg]$

$\sf \implies \dfrac{\sqrt{3} \pi^{2} }{18}$

$\sf \implies \displaystyle\int\limits^\frac{\pi}{4} _\frac{-\pi}{4} {\dfrac{x + \dfrac{\pi}{4} }{2 - cos2x} } \, dx = \dfrac{\sqrt{3} \pi^{2} }{18}$