Question

`  ∫    {1} / {cos x  + "cosec x" } dx  `

Answer 1

` \text{ Let  I }= ∫    {1} / {cos x  + "cosec x" } dx  `
`  ∫    {sin x} / {1 +cos x  + sin x } dx  ` 

`  ∫    {2 sin x} / { 2 + 2 cos  x  +sin x  } dx  `

 

`  ∫  {sin x + cos x + sin x - cos x }/ {2 + 2 cos x  sin x } dx  `

` ∫  {sin x + cos x  }/ {2 + 2 cos x  sin x } dx + ∫ {sin x - cos x} /{2 + 2 cos x  sin x } dx `

` ∫  {sin x + cos x  }/ {3 - sin^2 x -cos^2x+2 cos  x sin x} dx + ∫ {sin x - cos x} /{1 + sin^2 x + cos^2 x + 2cos x sin x } dx `
` ∫  {sin x + cos x  }/ {3 -( sin x -cos x )^2} dx + ∫ {sin x - cos x} /{1 + (sin x+ cos x + )^2} dx `
 where,` I_1 =  ∫  {sin x + cos x  }/ {3 -( sin x -cos x )^2} dx    and       I_2  =∫ {sin x - cos x} /{1 + (sin x+ cos x + )^2} dx `
\[Now, \]

` I_1 =  ∫  {sin x + cos x  }/ {3 -( sin x -cos x )^2} dx  `
` Let  ( sin x -cos x ) `
On differentiating both sides, we get

`(  cos x + sin x )dx = dt `
\[ \therefore I_1 = \int\frac{1}{3 - \left( t \right)^2}dt\]
\[ = \frac{1}{2\sqrt{3}}\text{ log }\left| \frac{\sqrt{3} + t}{\sqrt{3} - t} \right| + c_1 \]
\[ = \frac{1}{2\sqrt{3}}\text{ log }\left| \frac{\sqrt{3} + \sin x - \cos x}{\sqrt{3} - \left( \sin x - \cos x \right)} \right| + c_1 . . . (2)\]
\[Now, \]
\[ I_2 = \int\frac{\sin x - \cos x}{1 + \left( \sin x + \cos x \right)^2}dx\]
\[ Let \left( \sin x + \cos x \right) = t\]
 On  differentiating bothsides, weget
\[ \left( \cos x - \sin x \right)dx = dt\]
\[ \therefore I_2 = - \int\frac{1}{1 + \left( t \right)^2}dt\]
\[ = - \tan^{- 1} t + c_2 \]
\[ = - \tan^{- 1} \left( \sin x + \cos x \right) + c_2 . . . (3)\]
` "On substituting (2) and (3) in (1), we get" `
\[I = \frac{1}{2\sqrt{3}}\text{ log }\left| \frac{\sqrt{3} + \sin x - \cos x}{\sqrt{3} - \sin x + \cos x} \right| - \ tan^{- 1} \left( \sin x + \cos x \right) + c\]
\[Hence, \int\frac{1}{\cos x + cosec\ x}dx = \frac{1}{2\sqrt{3}}\text{ log }\left| \frac{\sqrt{3} + \sin x - \cos x}{\sqrt{3} - \sin x + \cos\ x} \right| - \tan^{- 1} \left( \sin x + \ cos\ x \right) + c\]