Question

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Answer 1

$\large\underline{\bold{Given \:Question - }}$

Evaluate the following integral :-

$\rm \: \displaystyle\int^{\pi} _{0} \sf \: \dfrac{x \: tanx }{secx \: cosecx} dx$

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

Identities Used :-

$\sf (1) \: \boxed{ \red{ \int\limits^a_0 \sf \: {f(x)} \, dx =\int\limits^a_0 {f(a - x)}}}$

$\sf (2) \: \boxed{ \red{ \int\limits^{2a}_0 \sf \: {f(x)} \, dx =2\int\limits^a_0 {f(x)} \: if \: f(a - x) = f(x)}}$

$(3). \: \boxed{ \red{ \sf \:sin(\pi \: - x) = sinx}}$

$(4). \: \boxed{ \red{ \sf \: {sin}^{2} x + {cos}^{2} x = 1}}$

$(5). \: \boxed{ \red{ \sf \: sin\bigg( \dfrac{\pi}{2} - x\bigg) = cosx}}$

Let's solve the problem now!!

$\rm :\longmapsto\:\rm \:I = \displaystyle\int^{\pi} _{0} \sf \: \dfrac{x \: tanx }{secx \: cosecx} dx$

$\rm \: I\: = \: \:\rm \: \displaystyle\int^{\pi} _{0} \sf \: \dfrac{x \: \dfrac{sinx}{cosx} }{\dfrac{1}{cosx} \times \dfrac{1}{sinx} } dx$

$\rm \: I \: = \: \:\rm \: \displaystyle\int^{\pi} _{0} \sf \:x\: {sin}^{2}x \:dx - - - (1)$

$\rm \: I \: = \: \:\rm \: \displaystyle\int^{\pi} _{0} \sf \:(\pi \: - x)\: {sin}^{2}(\pi \: - x) \:dx$

$\: \: \: \: \: \: \: \blue{\because \: \sf \int\limits^a_0 \bf \: {f(x)} \, dx =\int\limits^a_0 {f(a - x)}}$

$\rm \: I \: = \: \:\rm \: \displaystyle\int^{\pi} _{0} \sf \:(\pi \: - x)\: {sin}^{2}x \:dx - - - (2)$

On adding equation (1) and (2), we get

$\rm \: 2I \: = \: \:\rm \: \displaystyle\int^{\pi} _{0} \sf \:(x + \pi \: - x)\: {sin}^{2}x \:dx$

$\rm \: 2I \: = \: \:\rm \: \displaystyle\int^{\pi} _{0} \sf \pi\: {sin}^{2}x \:dx$

$\rm \: \cancel2I \: = \: \:\rm \: \cancel2 \displaystyle\int^{ \dfrac{\pi}{2} } _{0} \sf \pi\: {sin}^{2}x \:dx - - - (3)$

$\: \: \: \because \: \red{ \int\limits^{2a}_0 \sf \: {f(x)} \, dx =2\int\limits^a_0 {f(x)} \: as \: sin(\pi - x) = sinx}$

$\rm \: I \: = \: \:\rm \: \displaystyle\int^{ \dfrac{\pi}{2} } _{0} \sf \pi\: {sin}^{2}\bigg(\dfrac{\pi}{2} - x \bigg)\:dx$

$\rm \: I \: = \: \:\rm \: \displaystyle\int^{ \dfrac{\pi}{2} } _{0} \sf \pi\: {cos}^{2}x\:dx - - - (4)$

On adding equation (3) and equation (4), we get

$\rm \: 2I \: = \: \:\rm \: \displaystyle\int^{ \dfrac{\pi}{2} } _{0} \sf \pi\:( {sin}^{2}x + {cos}^{2}x)\:dx$

$\rm \: 2I \: = \: \:\rm \: \displaystyle\int^{ \dfrac{\pi}{2} } _{0} \sf \pi\:\:dx$

$\rm \:2I \: = \: \:\rm \:\pi \: \bigg( x\bigg)_0^{\dfrac{\pi}{2}}$

$\rm \:2I \: = \: \:\rm \:\pi \: \bigg(\dfrac{\pi}{2} - 0 \bigg)$

$\rm \:2I \: = \: \:\rm \:\pi \: \bigg(\dfrac{\pi}{2} \bigg)$

$\rm \:I \: = \: \:\rm \:\dfrac{ {\pi}^{2} }{4}$

Additional Information :-

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

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

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

$\sf (4) \: \int\limits^b_a {f(x)} \, dx = \int\limits^b_a{f(a + b - x)}$