Question

Solve it please
$\huge \bold\red{ \mathcal{\int {1 \div 1 + \cos(x.)dx}}}$
​

Answer 1

Answer:

we have to differentiate f(x)=\frac{1}{1-cosx}f(x)=

1−cosx

1

right ?

can we write f(x) = 1/(1 - cosx) = (1- cosx)^-1

now, using concepts : if a function y = f(x)^n is given then, differentiation of y = dy/dx = n . f(x)^(n - 1) × df(x)/dx

and also you should know differentiation of cosx = -sinx

now, differentiating f(x) = 1/(1 - cosx) or (1 - cosx)^-1

df(x)/dx = d{(1 - cosx)^-1}/dx

= -1(1 - cosx)^{-1 - 1} × d(1 - cosx)/dx

= -1(1 - cosx)^-2 × (0 -(-sinx))

= -1/(1 - cosx)² × sinx

= -sinx/(1 - cosx)²

hence, differentiation of 1/(1 - cosx) = -sinx/(1 - cosx)²

Answer 2

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

$\rm :\longmapsto\:\displaystyle\int\tt \dfrac{dx}{1 + cosx}$

We know,

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

Using this identity, we get

$\rm \: \: = \: \displaystyle\int\tt \dfrac{dx}{ {2cos}^{2}\dfrac{x}{2} }$

$\rm \: \: = \: \dfrac{1}{2}\displaystyle\int\tt {sec}^{2}\dfrac{x}{2} \: dx$

We know,

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

$\rm \: \: = \: \dfrac{1}{2} \times \dfrac{tan\dfrac{x}{2} }{\dfrac{1}{2} } + c$

$\rm \: \: = \: tan\dfrac{x}{2} + c$

Hence,

$\rm :\longmapsto\:\displaystyle\int\tt \dfrac{dx}{1 + cosx} = tan\dfrac{x}{2} \: + \: c$

Aliter Method :-

$\rm :\longmapsto\:\displaystyle\int\tt \dfrac{dx}{1 + cosx}$

On rationalizing the denominator, we get

$\rm \: \: = \: \:\displaystyle\int\tt \dfrac{1}{1 + cosx} \times \dfrac{1 - cosx}{1 - cosx} dx$

$\rm \: \: = \: \:\displaystyle\int\tt \dfrac{1 - cosx}{ {1}^{2} - cos^{2} x} dx$

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

$\rm \: \: = \: \displaystyle\int\tt \dfrac{1 - cosx}{ {sin}^{2}x } dx$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: {sin}^{2}x + {cos}^{2}x = 1 \bigg \}}$

$\rm \: \: = \: \displaystyle\int\tt \dfrac{1}{ {sin}^{2}x }dx - \displaystyle\int\tt \dfrac{cosx}{ {sin}^{2} x}dx$

$\rm \: \: = \: \displaystyle\int\tt {cosec}^{2}x \: dx - \displaystyle\int\tt \: cosecx \: cotx \: dx$

$\rm \: \: = \: - cotx \: + \: cosecx \: + c$

Hence,

$\bf :\longmapsto\:\displaystyle\int\tt \dfrac{dx}{1 + cosx} = cosecx \: - \: cotx \: + \: c$

Aliter Method :-

$\rm :\longmapsto\:\displaystyle\int\tt \dfrac{dx}{1 + cosx}$

We know,

$\red{ \boxed{ \sf{ \:cos2x = \frac{1 - {tan}^{2}x }{1 + {tan}^{2}x}}}}$

So, given integral can be reduced to

$\rm \: \: = \: \displaystyle\int\tt \dfrac{1}{1 - \dfrac{1 - {tan}^{2}\dfrac{x}{2} }{1 + {tan}^{2} \dfrac{x}{2} } } dx$

$\rm \: \: = \: \displaystyle\int\tt \dfrac{1}{\dfrac{1 + {tan}^{2}\dfrac{x}{2} + 1 - {tan}^{2}\dfrac{x}{2} }{1 + {tan}^{2} \dfrac{x}{2} } } dx$

$\rm \: \: = \: \displaystyle\int\tt \dfrac{1 + {tan}^{2} \dfrac{x}{2} }{2} \: dx$

$\rm \: \: = \: \dfrac{1}{2}\displaystyle\int\tt {sec}^{2}\dfrac{x}{2} \: dx$

$\red{ \boxed{ \sf{ \because \: {sec}^{2}x - {tan}^{2}x = 1 }}}$

$\rm \: \: = \: \dfrac{1}{2} \: \times \dfrac{tan\dfrac{x}{2} }{\dfrac{1}{2} } + c$

$\rm \: \: = \: tan\dfrac{x}{2} + c$

Hence,

$\rm :\longmapsto\:\displaystyle\int\tt \dfrac{dx}{1 + cosx} = tan\dfrac{x}{2} \: + \: c$

Formula Used

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

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

$\red{ \boxed{ \sf{\displaystyle\int\tt cosecx \: cotx \: dx \: = \: - \: cosecx + c}}}$

Additional Information :-

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

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

$\red{ \boxed{ \sf{\displaystyle\int\tt tanx \: dx \: = \:log secx + c }}}$

$\red{ \boxed{ \sf{\displaystyle\int\tt cotx \: dx \: = \:log sinx + c }}}$

$\red{ \boxed{ \sf{\displaystyle\int\tt cosecx \: dx \: = \:log (cosecx - cotx) + c }}}$

$\red{ \boxed{ \sf{\displaystyle\int\tt secx \: dx \: = \:log (secx + tanx) + c }}}$