Question

The anti-derivative of cos5x +cos4x/1-2cos3x​

Answer 1

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

The given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \dfrac{cos5x + cos4x}{1 - 2cos3x} \: dx$

The identities used to solve this integral are as follow,

$\boxed{ \sf{ \: cosx + cosy = 2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)}}$

$\boxed{ \sf{2sinxcosx = sin2x}}$

$\boxed{ \sf{ \: sinx - siny = 2cos\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{x - y}{2} \bigg)}}$

$\boxed{ \sf{2cosxcosy = cos(x + y) + cos(x - y)}}$

$\boxed{ \sf{\displaystyle\int\rm cosx = sinx + c}}$

Let's evaluate the problem now!!!

$\rm :\longmapsto\:\displaystyle\int\rm \dfrac{cos5x + cos4x}{1 - 2cos3x} \: dx$

$\rm \: = \: \displaystyle\int\rm \dfrac{2cos\bigg(\dfrac{5x + 4x}{2} \bigg)cos\bigg(\dfrac{5x - 4x}{2} \bigg)}{1 - 2cos3x} dx$

On multiply and divide by sin3x, we get

$\rm \: = \: \displaystyle\int\rm \dfrac{sin3x \times \: 2cos\bigg(\dfrac{9x}{2} \bigg)cos\bigg(\dfrac{x}{2} \bigg)}{sin3x(1 - 2cos3x)} dx$

$\rm \: = \:\displaystyle\int\rm \dfrac{2sin\bigg(\dfrac{3x}{2} \bigg)cos\bigg(\dfrac{3x}{2} \bigg) \times \: 2cos\bigg(\dfrac{9x}{2} \bigg)cos\bigg(\dfrac{x}{2} \bigg)}{sin3x- 2sin3xcos3x}dx$

$\rm \: = \:\displaystyle\int\rm \dfrac{4sin\bigg(\dfrac{3x}{2} \bigg)cos\bigg(\dfrac{3x}{2} \bigg) cos\bigg(\dfrac{9x}{2} \bigg)cos\bigg(\dfrac{x}{2} \bigg)}{sin3x- sin6x}dx$

$\rm \: = \:\displaystyle\int\rm \dfrac{4sin\bigg(\dfrac{3x}{2} \bigg)cos\bigg(\dfrac{3x}{2} \bigg) cos\bigg(\dfrac{9x}{2} \bigg)cos\bigg(\dfrac{x}{2} \bigg)}{2cos\bigg(\dfrac{3x + 6x}{2} \bigg)sin\bigg(\dfrac{3x - 6x}{2} \bigg)}dx$

$\rm \: = \:\displaystyle\int\rm \dfrac{2sin\bigg(\dfrac{3x}{2} \bigg)cos\bigg(\dfrac{3x}{2} \bigg) \cancel{cos\bigg(\dfrac{9x}{2} \bigg)}cos\bigg(\dfrac{x}{2} \bigg)}{ \cancel{cos\bigg(\dfrac{9x}{2} \bigg)} \: sin\bigg(\dfrac{- 3x}{2} \bigg)}dx$

$\rm \: = \:\displaystyle\int\rm \dfrac{2 \: \cancel{sin\bigg(\dfrac{3x}{2} \bigg)} \: cos\bigg(\dfrac{3x}{2} \bigg) cos\bigg(\dfrac{x}{2} \bigg)}{ \: - \: \cancel{sin\bigg(\dfrac{3x}{2} \bigg)}}dx$

$\rm \: = \: - \displaystyle\int\rm \: 2 cos\bigg(\dfrac{3x}{2} \bigg)cos\bigg(\dfrac{x}{2} \bigg) \: dx$

$\rm \: = \: - \displaystyle\int\rm \bigg \{cos\bigg(\dfrac{3x}{2} + \dfrac{x}{2} \bigg) + cos\bigg(\dfrac{3x}{2} - \dfrac{x}{2} \bigg) \bigg \}dx$

$\rm \: = \: - \displaystyle\int\rm (cos2x + cosx)dx$

$\rm \: = \: - \: \dfrac{sin2x}{2} + sinx + c$

Hence,

$\purple{\boxed{ \bf{ \: \displaystyle\int\bf \dfrac{cos5x + cos4x}{1 - 2cos3x}dx\:=-\: \dfrac{sin2x}{2} + sinx + c}}}$

Additional Information :-

$\boxed{ \sf{\displaystyle\int\rm sinx = - cosx + c}}$

$\boxed{ \sf{\displaystyle\int\rm tanx = - logcosx + c}}$

$\boxed{ \sf{\displaystyle\int\rm tanx = logsecx + c}}$

$\boxed{ \sf{\displaystyle\int\rm cotx = logsinx + c}}$

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

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

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

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