Math, asked by INSIDI0US, 4 months ago

Evaluate the following ! :)

 \Large {\displaystyle \sf \int \limits^{\dfrac{\pi}{4}}_ {\dfrac{- \pi}{4}} \small \dfrac{x\ +\ \dfrac{\pi}{4}}{2\ -\ cos\ 2x} dx}

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Answered by mathdude500
75

\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}

We know,

\begin{gathered}\begin{gathered}\bf\: \displaystyle \sf \int \limits^{a}_ { - a} \: f(x) \:  dx = \begin{cases} &\sf{0, \:  \: if \: f( - x) =  - f(x)} \\ &\sf{2\displaystyle \sf \int \limits^{a}_ {0} \: f(x) \:  dx, \:  \: if \: f( - x) = f(x)} \end{cases}\end{gathered}\end{gathered}

\large\underline{\sf{Solution-}}

\rm :\longmapsto\:Let \: I \:  =  \: \displaystyle \sf \int \limits^{\dfrac{\pi}{4}}_ {\dfrac{- \pi}{4}} \small \dfrac{x\ +\ \dfrac{\pi}{4}}{2\ -\ cos\ 2x} dx

\:  \rm \:I =  \: \: \displaystyle \sf \int \limits^{\dfrac{\pi}{4}}_ {\dfrac{- \pi}{4}} \small \dfrac{x}{2\ -\ cos\ 2x} dx  \:  +  \: \: \displaystyle \sf \int \limits^{\dfrac{\pi}{4}}_ {\dfrac{- \pi}{4}} \small \dfrac{\ \dfrac{\pi}{4}}{2\ -\ cos\ 2x} dx

\:  \rm\:Let\: I  \: =  \:  \:\: \displaystyle \sf \int \limits^{\dfrac{\pi}{4}}_ {\dfrac{- \pi}{4}} g(x) + \: \displaystyle \sf \int \limits^{\dfrac{\pi}{4}}_ {\dfrac{- \pi}{4}}h(x) -  -  - (1)

where,

\rm :\longmapsto\:g(x) =\dfrac{x}{2\ -\ cos\ 2x}

and

\rm :\longmapsto\:h(x) = \dfrac{\ \dfrac{\pi}{4}}{ \:  \: 2\ -\ cos\ 2x \:  \:  \: }

Now,

Consider,

\rm :\longmapsto\:g( - x)

\:  \rm\:  \:  =  \: \dfrac{ - x}{2\ -\ cos\ 2( - x)}

\:  \rm\:  \:  =  \: \dfrac{ - x}{2\ -\ cos\ 2x} \:  \:  \: \:  \:   \{ \because \: cos( - x) = cosx \}

\:  \rm \:  \:  =  \:  \:  - g(x)

\boxed{\red{\sf\: \therefore \: \: \displaystyle \sf \int \limits^{\dfrac{\pi}{4}}_ {\dfrac{- \pi}{4}}g(x) = 0}} -  -  - (2)

Now,

Consider,

\rm :\longmapsto\:h( - x)

\rm :\longmapsto\:h( - x) = \dfrac{\ \dfrac{\pi}{4}}{ \:  \: 2\ -\ cos\ 2( - x) \:  \:  \: }

\rm :\longmapsto\:h( - x) = \dfrac{\ \dfrac{\pi}{4}}{ \:  \: 2\ -\ cos\ 2x \:  \:  \: }

\rm :\implies\:h( - x) = h(x)

\rm :\implies\:\: \displaystyle \sf \int \limits^{\dfrac{\pi}{4}}_ {\dfrac{- \pi}{4}} \small \dfrac{\ \dfrac{\pi}{4}}{2\ -\ cos\ 2x} dx =2 \: \displaystyle \sf \int \limits^{\dfrac{\pi}{4}}_ {0} \small \dfrac{\ \dfrac{\pi}{4}}{2\ -\ cos\ 2x} dx

\:  \rm \:  \:  =  \:  \: 2 \: \displaystyle \sf \int \limits^{\dfrac{\pi}{4}}_ {0} \small \dfrac{\ \dfrac{\pi}{4}}{2\ - \: \dfrac{1 -  {tan}^{2}x }{1 +  {tan}^{2}x } } dx

\:  \rm \:  \:  =  \:  \: 2 \times \dfrac{\pi}{4}  \: \displaystyle \sf \int \limits^{\dfrac{\pi}{4}}_ {0} \small \dfrac{1}{\: \dfrac{2 + 2 {tan}^{2}x -  1  +  {tan}^{2}x }{1 +  {tan}^{2}x } } dx

\:  \rm \:  \:  =  \:  \dfrac{\pi}{2}  \: \displaystyle \sf \int \limits^{\dfrac{\pi}{4}}_ {0} \small \dfrac{1 +  {tan}^{2} x}{\: 3 {tan}^{2}x  + 1} dx

\:  \rm \:  \:  =  \:  \dfrac{\pi}{2}  \: \displaystyle \sf \int \limits^{\dfrac{\pi}{4}}_ {0} \small \dfrac{{sec}^{2} x}{\: 3 {tan}^{2}x  + 1} dx

\:  \rm \:  \:  =  \:  \dfrac{\pi}{6}  \: \displaystyle \sf \int \limits^{\dfrac{\pi}{4}}_ {0} \small \dfrac{{sec}^{2} x}{\:  {tan}^{2}x  + \dfrac{1}{3} } dx

 \bigg \{ \sf \: Put \: tanx = y \\  \sf \:  {sec}^{2}x \: dx = dy  \\  \sf \: As \: x \to0, \: y \to \: 0 \\  \sf \: As \: x \to \: \dfrac{\pi}{4}, \: y \to1  \bigg\}

\:  \rm \:  \:  =  \:  \dfrac{\pi}{6}  \: \displaystyle \sf \int \limits^{1}_ {0} \small \dfrac{1}{\:  {y}^{2}  + \dfrac{1}{3} } dy

\:  \rm \:  \:  =  \:  \dfrac{\pi}{6}  \: \displaystyle \sf \int \limits^{1}_ {0} \small \dfrac{1}{\:  {y}^{2}  + \dfrac{1}{ {( \sqrt{3}) }^{2} } } dy

 \:  \:  \rm \:   =  \dfrac{ \sqrt{3} \pi}{6} \bigg(  {tan}^{ - 1} \sqrt{3}y  \bigg)_0^1

 \:  \:  \rm \:   =  \dfrac{ \sqrt{3} \pi}{6} \bigg(  {tan}^{ - 1} \sqrt{3} -  {tan}^{ - 1}0   \bigg)

 \:  \:  \rm \:   =  \dfrac{ \sqrt{3} \pi}{6}  \times \dfrac{\pi}{3}

 \:  \:  \rm \:   =  \dfrac{ \sqrt{3} \pi ^{2} }{18}

\rm :\implies\:\boxed{\red{\sf\:\: \displaystyle \sf \int \limits^{\dfrac{\pi}{4}}_ {\dfrac{- \pi}{4}}h(x) = \dfrac{ \sqrt{3} {\pi}^{2}  }{18} }} -  -  - (3)

On substituting the values from equation (2) and (3) in equation (1), we get

\rm :\implies\:I = 0 + \dfrac{ \sqrt{3} {\pi}^{2}  }{18}

\rm :\implies\:I = \dfrac{ \sqrt{3} {\pi}^{2}  }{18}

 \boxed{\purple{\bf\implies \:{\displaystyle \sf \int \limits^{\dfrac{\pi}{4}}_ {\dfrac{- \pi}{4}} \small \dfrac{x\ +\ \dfrac{\pi}{4}}{2\ -\ cos\ 2x} dx} = \dfrac{ \sqrt{3}  {\pi}^{2} }{18}}}

Additional Information :-

\sf (1) \:  \int\limits^b_a {f(x)} \, dx = \int\limits^b_a {f(t)}

\sf (2) \: \int\limits^b_a {f(x)} \, dx = -\int\limits^a_b {f(x)}

\sf (3) \:  \int\limits^b_a {f(x)} \, dx = \int\limits^c_a {f(x)} \, dx + \int\limits^b_c {f(x)} \, dx \ where\ \ a < c < b

\sf (4) \:  \int\limits^a_0 {f(x)} \, dx =\int\limits^a_0 {f(a - x)}

\sf (5) \: \int\limits^b_a {f(x)} \, dx = \int\limits^b_a{f(a + b - x)}

Answered by xxavneetxx
3

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