Question

The integral $\int {x{{\cos }^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)dx} $ is equal to
A) $ - x + \left( {1 + {x^2}} \right){\cot ^{ - 1}}x + c$
B) $ - x + \left( {1 + {x^2}} \right){\tan ^{ - 1}}x + c$
C) $ - x - \left( {1 + {x^2}} \right){\tan ^{ - 1}}x + c$
D) $ - x - \left( {1 + {x^2}} \right){\cot ^{ - 1}}x + c$

Answer 1

Hint:
In order to find the given integral we need to substitute $x = \tan \theta $,then differentiating it we get $dx = {\sec ^2}\theta d\theta $.and then by using identity $\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \cos 2\theta $ and ${\cos ^{ - 1}}\left( {\cos \theta } \right) = \theta $we get the integral $\int {2\theta \tan \theta s} e{c^2}\theta d\theta $and we need to integrate this by using $\int {uvdx = u\int {vdx} - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} dx} $ ,where $u = 2\theta $ and $v = \tan \theta {\sec ^2}\theta $ and further simplification is done by using the identity ${\sec ^2}\theta - {\tan ^2}\theta = 1$and finally replacing by x we get the required answer.

Complete step by step solution:
We are asked to find the integral of $\int {x{{\cos }^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)dx} $
So lets start by using substitution method
Lets take $x = \tan \theta $
Then differentiating it we get $dx = {\sec ^2}\theta d\theta $
$ \Rightarrow \int {\tan \theta {{\cos }^{ - 1}}\left( {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)s} e{c^2}\theta d\theta $
Now we know the identity $\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \cos 2\theta $
Using this in the above equation we get
 $ \Rightarrow \int {\tan \theta {{\cos }^{ - 1}}\left( {\cos 2\theta } \right)s} e{c^2}\theta d\theta $
And we know that ${\cos ^{ - 1}}\left( {\cos \theta } \right) = \theta $
Using this we get
$ \Rightarrow \int {2\theta \tan \theta s} e{c^2}\theta d\theta $
Now here we can do integration by parts
$\int {uvdx = u\int {vdx} - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} dx} $
Here $u = 2\theta $ and $v = \tan \theta {\sec ^2}\theta $
Hence we get
\[
   \Rightarrow \int {2\theta \tan \theta s} e{c^2}\theta d\theta = 2\theta \int {\tan \theta {{\sec }^2}\theta d\theta } - \int {\left( {\dfrac{{d(2\theta )}}{{d\theta }}\int {\tan \theta {{\sec }^2}\theta d\theta } } \right)} d\theta \\
   \Rightarrow \int {2\theta \tan \theta s} e{c^2}\theta d\theta = 2\theta \int {\tan \theta {{\sec }^2}\theta d\theta } - \int {\left( {2\int {\tan \theta {{\sec }^2}\theta d\theta } } \right)} d\theta \\
 \]
Now once again we need to use substitution method to solve the integrals
Lets take $t = \tan \theta $
Then differentiating it we get $dt = {\sec ^2}\theta d\theta $
\[
   \Rightarrow \int {2\theta \tan \theta s} e{c^2}\theta d\theta = 2\theta \int {\operatorname{t} dt} - \int {\left( {2\int {tdt} } \right)} \dfrac{{dt}}{{{{\sec }^2}\theta }} \\
   \Rightarrow \int {2\theta \tan \theta s} e{c^2}\theta d\theta = 2\theta \dfrac{{{t^2}}}{2} - \int {\left( {2\dfrac{{{t^2}}}{2}} \right)} \dfrac{{dt}}{{{{\sec }^2}\theta }} \\
   \Rightarrow \int {2\theta \tan \theta s} e{c^2}\theta d\theta = \theta {t^2} - \int {\left( {{t^2}} \right)} \dfrac{{dt}}{{{{\sec }^2}\theta }} \\
 \]
Now replacing t by $\tan \theta $
\[ \Rightarrow \int {2\theta \tan \theta s} e{c^2}\theta d\theta = \theta {\tan ^2}\theta - \int {\left( {{{\tan }^2}\theta } \right)} d\theta \]
We know that by using the identity ${\sec ^2}\theta - {\tan ^2}\theta = 1$
We can write it as ${\sec ^2}\theta - 1 = {\tan ^2}\theta $
Using this we get
\[
   \Rightarrow \int {2\theta \tan \theta s} e{c^2}\theta d\theta = \theta {\tan ^2}\theta - \int {\left( {{{\sec }^2}\theta - 1} \right)} d\theta \\
   \Rightarrow \int {2\theta \tan \theta s} e{c^2}\theta d\theta = \theta {\tan ^2}\theta - \left[ {\int {{{\sec }^2}\theta } d\theta - \int {d\theta } } \right] \\
   \Rightarrow \int {2\theta \tan \theta s} e{c^2}\theta d\theta = \theta {\tan ^2}\theta - \left[ {\tan \theta - \theta } \right] \\
   \Rightarrow \int {2\theta \tan \theta s} e{c^2}\theta d\theta = \theta {\tan ^2}\theta - \tan \theta + \theta \\
   \Rightarrow \int {2\theta \tan \theta s} e{c^2}\theta d\theta = \theta \left[ {1 + {{\tan }^2}\theta } \right] - \tan \theta \\
 \]
Now replacing $\theta $ by x we get
\[ \Rightarrow \int {2\theta \tan \theta s} e{c^2}\theta d\theta = {\tan ^{ - 1}}x\left[ {1 + {x^2}} \right] - x\]
Hence we get
$ \Rightarrow \int {x{{\cos }^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)dx} = - x + \left( {1 + {x^2}} \right){\tan ^{ - 1}}x + c$

Therefore the correct option is B.

Note:
Steps to keep in mind while solving trigonometric problems are
1) Always start from the more complex side
2) Express everything into sine and cosine
3) Combine terms into a single fraction
4) Use Pythagorean identities to transform between ${\sin ^2}x{\text{ and }}{\cos ^2}x$
5) Know when to apply double angle formula
6) Know when to apply addition formula
7) Good old expand/ factorize/ simplify/ cancelling
8) Integration, in mathematics, technique of finding a function g(x) the derivative of which, Dg(x), is equal to a given function f(x). This is indicated by the integral sign $\int {} $ as in $\int {f(x)} $ , usually called the indefinite integral of the function.