Question

Integration of (2cos x + 1)^2 dx
Please solve this step by step. Thanks

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm {(2cosx + 1)}^{2} \: dx$

$\rm \: = \: \: \:\displaystyle\int\rm ({4cos}^{2}x + 1 + 4cosx) \: dx$

$\red{\bigg \{ \because \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy\bigg \}}$

$\rm \: = \: \: \:\displaystyle\int\rm (2{(2cos}^{2}x )+ 1 + 4cosx) \: dx$

We know,

$\boxed{ \bf{ \: 1 + cos2x = {2cos}^{2}x }}$

$\rm \: = \: \: \:\displaystyle\int\rm (2(1 + cos2x)+ 1 + 4cosx) \: dx$

$\rm \: = \: \: \:\displaystyle\int\rm (2 + 2cos2x+ 1 + 4cosx) \: dx$

$\rm \: = \: \: \:\displaystyle\int\rm (2cos2x+ 3 + 4cosx) \: dx$

We know,

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

and

$\boxed{ \bf{ \displaystyle\int\rm kdx \: = \: \: kx + c }}$

$\rm \: = \: \: 2 \times \dfrac{sin2x}{2} + 3x + 4sinx + c$

$\rm \: = \: sin2x+ 3x + 4sinx + c$

Hence,

$\boxed{ \bf{\displaystyle\int\bf {(2cosx + 1)}^{2} \: dx =sin2x + 4sinx + 3x + c }}$

Additional Information :-

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

$\boxed{ \bf{\displaystyle\int\rm tanx \: dx \: = \: log \:( secx )\: + \: c }}$

$\boxed{ \bf{\displaystyle\int\rm cotx \: dx \: = \: log \:( sinx )\: + \: c }}$

$\boxed{ \bf{\displaystyle\int\rm secx \: dx \: = \: log \:( secx + tanx)\: + \: c }}$

$\boxed{ \bf{\displaystyle\int\rm cosecx \: dx \: = \: log \:( cosecx - cotx)\: + \: c }}$

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

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