Solve $ \int {\left[ {d\dfrac{{(\cos \theta )}}{{(\sqrt {1 - {{\cos }^2}\theta )} }}} \right]} = $
$
  1){\cos ^{ - 1}}\theta + c \\
  2)\theta + c \\
  3){\sin ^{ - 1}}\theta + c \\
  4){\sin ^{ - 1}}(\cos \theta ) + c \;
  $

Question

Solve $ \int {\left[ {d\dfrac{{(\cos \theta )}}{{(\sqrt {1 - {{\cos }^2}\theta )} }}} \right]} = $
$
  1){\cos ^{ - 1}}\theta + c \\
  2)\theta + c \\
  3){\sin ^{ - 1}}\theta + c \\
  4){\sin ^{ - 1}}(\cos \theta ) + c \;
  $

Vedantu Content Team · Accepted Answer

Hint:  First of all we will convert the given expression in the simplified form and then place the relevant formulas and simplify for the resultant required values. To simplify the expression, take any variable as the reference term and replace it with the cosine function.Complete step-by-step answer:Let us assume that, $ \cos \theta  = y $  			…. (A)Take derivative on both the sides of the equation – $ \dfrac{d}{{d\theta }}(\cos \theta ) = \dfrac{d}{{d\theta }}(y) $ Use the formula of derivative on the left hand side of the above equation – $  - \sin \theta  = \dfrac{{dy}}{{d\theta }} $ Perform cross multiplication in the above expression while the numerator of one side is multiplied with the denominator of the opposite side. $  - \sin \theta d\theta  = dy $ Take the given expression: $ \int {\left[ {d\dfrac{{(\cos \theta )}}{{(\sqrt {1 - {{\cos }^2}\theta )} }}} \right]}  $ Use the value for equation (A) in the above expression – $  = \int {\left[ {d\dfrac{y}{{(\sqrt {1 - {y^2})} }}} \right]}  $ The above expression can be re-written as – $  = \int {\left[ {\dfrac{{dy}}{{(\sqrt {1 - {y^2})} }}} \right]}  $ Now using the identity – the above expression can be replaced as – $  = {\sin ^{ - 1}}y + c $ Now again using the value of the equation (A) $  = {\sin ^{ - 1}}(\cos \theta ) + c $ From the given multiple choices, the fourth option is the correct answer.So, the correct answer is “Option 4”.Note: Always remember that the inverse of the cosine function with the cosine function cancels each other. Same trigonometric function and its inverse cancel each other out. Always remember the formulas for the integration of trigonometric functions and its value for efficiency and the correct solution and also be careful about the sign convention. However if we substitute  $  - \sin \theta d\theta  = dy $ the equation becomes  $ -\theta + c$ but that's not in the option.

Solve $ \int {\left[ {d\dfrac{{(\cos \theta )}}{{(\sqrt {1 - {{\cos }^2}\theta )} }}} \right]} = $ $ 1){\cos ^{ - 1}}\theta + c \\ 2)\theta + c \\ 3){\sin ^{ - 1}}\theta + c \\ 4){\sin ^{ - 1}}(\cos \theta ) + c \; $

Solve $ \int {\left[ {d\dfrac{{(\cos \theta )}}{{(\sqrt {1 - {{\cos }^2}\theta )} }}} \right]} = $
$
1){\cos ^{ - 1}}\theta + c \\
2)\theta + c \\
3){\sin ^{ - 1}}\theta + c \\
4){\sin ^{ - 1}}(\cos \theta ) + c \;
$