Question

Can anyone help me with this( preferably with steps) ​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int_0^{ \dfrac{\pi}{2} } \: \frac{ {sin}^{2} x}{ {(1 + cosx)}^{2} } \: dx$

We know,

$\red{\boxed{ \rm{ {sin}^{2}x + {cos}^{2}x = 1}}}$

So, using this, we can rewrite the above integral as

$\rm \: = \: \:\displaystyle\int_0^{ \dfrac{\pi}{2} } \: \frac{1 - {cos}^{2} x}{ {(1 + cosx)}^{2} } \: dx$

We know,

$\red{\boxed{ \rm{ {x}^{2} - {y}^{2} = (x + y)(x - y)}}}$

$\rm \: = \: \:\displaystyle\int_0^{ \dfrac{\pi}{2} } \: \frac{(1 - {cos}x)(1 + cosx)}{ {(1 + cosx)}^{2} } \: dx$

$\rm \: = \: \:\displaystyle\int_0^{ \dfrac{\pi}{2} } \: \frac{1 - {cos}x}{ {1 + cosx}} \: dx$

We know,

$\red{\boxed{ \rm{1 - cos2x = {2sin}^{2}x}}} \: \: and \: \: \red{\boxed{ \rm{1 + cos2x = {2sin}^{2}x}}}$

So, using these Identities, we get

$\rm \: = \: \:\displaystyle\int_0^{ \dfrac{\pi}{2} } \: \frac{2 {sin}^{2}\dfrac{x}{2} }{ {2cos}^{2}\dfrac{x}{2} } \: dx$

$\rm \: = \: \:\displaystyle\int_0^{ \dfrac{\pi}{2} } \: \frac{ {sin}^{2}\dfrac{x}{2} }{ {cos}^{2}\dfrac{x}{2} } \: dx$

$\rm \: = \: \:\displaystyle\int_0^{ \dfrac{\pi}{2} } \: \ {tan}^{2} \frac{x}{2} \: dx$

We know,

$\red{\boxed{ \rm{ {sec}^{2}x - {tan}^{2}x = 1}}}$

So, we get

$\rm \: = \: \:\displaystyle\int_0^{ \dfrac{\pi}{2} } \: (\ {sec}^{2} \frac{x}{2} - 1) \: dx$

Now, we know,

$\red{\boxed{ \rm{\displaystyle\int \: {sec}^{2}xdx = tanx +c}}} \: \: and \: \: \red{\boxed{ \rm{\displaystyle\int \: kdx = kx + c}}}$

So, using these we get

$\rm \: = \: \bigg(\dfrac{tan\dfrac{x}{2} }{\dfrac{1}{2} } - x\bigg)_0^{ \dfrac{\pi}{2} }$

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

$\rm \: = \: \bigg(2tan\dfrac{\pi}{4} - \dfrac{\pi}{2}\bigg) - (0 - 0)$

$\rm \: = \: 2 - \dfrac{\pi}{2}$

$\rm \: = \: \dfrac{4 - \pi}{2}$

Hence,

$\red{\rm :\longmapsto\:\displaystyle\int_0^{ \dfrac{\pi}{2} } \bf\: \frac{ {sin}^{2} x}{ {(1 + cosx)}^{2} } \: dx = \frac{4 - \pi}{2} }$

Additional Information :-

$\red{\boxed{ \rm{\displaystyle\int \: sinx \: dx = - cosx + c}}}$

$\red{\boxed{ \rm{\displaystyle\int \: cosx \: dx = sinx + c}}}$

$\red{\boxed{ \rm{\displaystyle\int \: {sec}^{2} x \: dx = tanx + c}}}$

$\red{\boxed{ \rm{\displaystyle\int \: {cosec}^{2} x \: dx= - cotx + c}}}$

$\red{\boxed{ \rm{\displaystyle\int \: {cosec}x \: cotx \: dx = - cosecx + c}}}$

$\red{\boxed{ \rm{\displaystyle\int \: secx \: tanx \: dx = secx + c}}}$