Question

#Integrals

#Quality Question >>>


Evaluate

$\int \dfrac{1}{ {cos}^{4}x + {sin}^{4}x } dx$


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Answer 1

$I= \int \dfrac{1}{ {cos}^{4}x + {sin}^{4}x } dx$

$\sf Divide \: Numerator \: and \: Denominator \\ \sf \: by \: {cos}^{4} x$

$I = \int\dfrac{ {sec}^{4}x}{1 + {tan}^{4}x } dx \\ \\ = \int \frac{ {sec}^{2} x . {sec}^{2}x }{1 + {tan}^{4}x } dx \\ \\ = \int \frac{1 + {tan}^{2} x}{1 + {tan}^{4}x } . {sec}^{2}x \: dx \\ \\ \sf put \: tanx = z\implies {sec}^{2} x \: dx = dz\\ \\ I = \int \frac{1 + {z}^{2} }{1 + {z}^{4} } \: dz \\ \\ = \int \dfrac{ \dfrac{1}{ {z}^{2} } + 1 }{ \dfrac{1}{ {z}^{2} } + {z}^{2} } \: dz \\ \\ = \int \dfrac{ \dfrac{1}{ {z}^{2} } + 1 }{( z - \dfrac{1}{ {z}^{} } {)}^{2} + 2} \: dz$

$\sf let \: z - \dfrac{1}{z} = t \implies (1 + \dfrac{1}{ {z}^{2} } )dz = dt$

$I = \int \dfrac{dt}{ {t}^{2} + ( \sqrt{ {2} } {)}^{2} } \\ \\ = \sf \frac{1}{ \sqrt{2} } {tan}^{ - 1} \frac{t}{ \sqrt{2} } + c \\ \\ = \sf \frac{1}{ \sqrt{2} } \: \: {tan}^{ - 1} \: \: \frac{z - \dfrac{1}{z} }{ \sqrt{2} } + c \\ \\ \sf \frac{1}{ \sqrt{2} } \: tan {}^{ - 1} \: \frac{ {z}^{2} - 1 }{ \sqrt{2} \: z} + c \\ \\ = \sf \frac{1}{ \sqrt{2} } \: tan {}^{ - 1} \: \frac{ {tan}^{2}x - 1 }{ \sqrt{2} \: tan \: x} + c$

Answer 2

EXPLANATION.

$\sf \implies \displaystyle \int \dfrac{1}{cos^{4}x + sin^{4}x } dx$

As we know that,

Divide numerator and denominator by cos⁴(x), we get.

$\sf \implies \displaystyle \int \dfrac{\dfrac{1}{cos^{4}x } }{\dfrac{cos^{4} x}{cos^{4}x }\ + \dfrac{sin^{4}x }{cos^{4} x} } \ dx$

$\sf \implies \displaystyle \int \dfrac{sec^{4} x \ dx}{1 + tan^{4}x }$

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

As we know that,

Formula of :

⇒ sec²x = 1 + tan²x.

Using this formula in equation, we get.

$\sf \implies \displaystyle \int \dfrac{(1 + tan^{2} x) sec^{2}x \ dx }{(1 + tan^{4}x) }$

By using the substitution method, we get.

Let we assume that,

⇒ tan x = t.

Differentiate w.r.t x, we get.

⇒ sec²x dx = dt.

Put the values in the equation, we get.

$\sf \implies \displaystyle \int \dfrac{(1 + t^{2})dt }{(1 + t^{4}) }$

Divide numerator and denominator by t², we get.

$\sf \implies \displaystyle \int \dfrac{\bigg(\dfrac{1 + t^{2} }{t^{2} } \bigg)}{\bigg(\dfrac{1 + t^{4} }{t^{2} } \bigg)} \ dt$

$\sf \implies \displaystyle \int \dfrac{\bigg(1 + \dfrac{1}{t^{2}} \bigg)}{\bigg(\dfrac{1}{t^{2} } + t^{2} \bigg)} dt$

$\sf \implies \displaystyle \int \dfrac{\bigg( 1 + \dfrac{1}{t^{2} } \bigg)}{\bigg( t - \dfrac{1}{t} \bigg)^{2} + 2} dt$

Again we apply substitution method, we get.

Let we assume that,

⇒ t - 1/t = z.

⇒ (1 + 1/t²)dt = dz.

Put the values in the equation, we get.

$\sf \implies \displaystyle \int \dfrac{dz}{z^{2} + 2} = \int \dfrac{dt}{z^{2} + (\sqrt{2} )^{2} }$

As we know that,

Formula of :

$\sf \implies \displaystyle \int \dfrac{1}{x^{2} + a^{2} } \ = \dfrac{1}{a} tan^{-1} \bigg(\frac{x}{a} \bigg) + C.$

Using this formula in equation, we get.

$\sf \implies \displaystyle \dfrac{1}{\sqrt{2} } tan^{-1} \bigg( \dfrac{z}{\sqrt{2} } \bigg) + C.$

Put the value of z = t - 1/t in equation, we get.

$\sf \implies \displaystyle \dfrac{1}{\sqrt{2} } tan^{-1} \bigg( \dfrac{t - \dfrac{1}{t} }{\sqrt{2} } \bigg) + C.$

$\sf \implies \displaystyle \dfrac{1}{\sqrt{2} } tan^{-1} \bigg( \dfrac{t^{2} - 1}{\sqrt{2}t } \bigg) + C.$

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

$\sf \implies \displaystyle \dfrac{1}{\sqrt{2} } tan^{-1} \bigg(\dfrac{tan^{2}x - 1 }{\sqrt{2}tan (x) } \bigg) + C.$

$\sf \implies \displaystyle \int \frac{1}{cos^{4} x + sin^{4} x} dx \ = \displaystyle \dfrac{1}{\sqrt{2} } tan^{-1} \bigg(\dfrac{tan^{2}x - 1 }{\sqrt{2}tan (x) } \bigg) + C.$

                                                                                                                       

MORE INFORMATION.

Important points.

If a function can be expressed in terms of elementary function (formula format) then only it is integrable, other wise cannot.

For example :

∫e^(sin x) dx , ∫√sin(x) dx , ∫x⁴/x¹⁰ + 1 dx , ∫dx/㏑ sin x dx.