Question

Evaluate $\int {{{\tan }^{ - 1}}} \sqrt x dx$.

Answer 1

Hint: we can start by using integration by parts. Remember, $u$and $v$are functions of $x$with $u'$and $v'$their respective derivatives with respect to $x$. Then:
$\int {uvdx = u\int {vdx - \int {(u'\int {vdx} } } } )dx$
Now, the important step is the selection of functions as $u$and $v$. For that, $ILATE$ rule can be used. $I$stands for inverse trigonometric functions, $L$for logarithmic functions, $A$ for algebraic functions, $T$ for trigonometric functions, and $E$for exponential functions.
Using this rule, we can fix $u$ and $v$. Then, using the above formula, we can integrate easily.

Complete Step by Step Solution:
Here, ${\tan ^{ - 1}}\sqrt x $ is an inverse trigonometric function. So, it can be taken as $u$. As we can see in the integration $\int {{{\tan }^{ - 1}}} \sqrt x dx$, there is no other function of $x$which can be considered as $v$.
$\int {{{\tan }^{ - 1}}} \sqrt x dx$ can be written as $\int {({{\tan }^{ - 1}}} \sqrt x \times 1)dx$.
$1$ can be considered as function of $x$. ($\because 1 = {x^0}$)
In $\int {{{\tan }^{ - 1}}} \sqrt x dx$, $u = {\tan ^{ - 1}}\sqrt x $and $v = 1$
Using by parts by formula,
$\int {uvdx = u\int {vdx - \int {(u'\int {vdx} } } } )dx$
Let $I = \int {{{\tan }^{ - 1}}} \sqrt x dx$
$\int {{{\tan }^{ - 1}}} \sqrt x dx = {\tan ^{ - 1}}\sqrt x \int {1dx} - \int {[\dfrac{d}{{dx}}} ({\tan ^{ - 1}}\sqrt x )\int {1dx]dx} .................................($equation $1$)
Integration of $1$is $x$.
Differentiation of ${\tan ^{ - 1}}x = \dfrac{1}{{1 + {x^2}}}$
Replace $x$with $\sqrt x $;
Differentiation of ${\tan ^{ - 1}}\sqrt x = \dfrac{1}{{1 + x}} \times \dfrac{1}{{2\sqrt x }}$ = ($\dfrac{1}{{2\sqrt x }}$ ,due to internal differentiation of $\sqrt x $)
Equation $1$ will become,
$I = x{\tan ^{ - 1}}\sqrt x - \int {\dfrac{x}{{1 + x}}} \times \dfrac{1}{{2\sqrt x }}dx$
$I = x{\tan ^{ - 1}}\sqrt x - \dfrac{1}{2}\int {\dfrac{{\sqrt x }}{{1 + x}}} dx........................$(equation $2$)
Let ${I_2} = \int {\dfrac{{\sqrt x }}{{1 + x}}} dx$
Let $\sqrt x = y$$ \Rightarrow x = {y^2}$
$ \Rightarrow dx = 2ydy$
${I_2} = \int {\dfrac{y}{{1 + {y^2}}}} \times 2ydy$
${I_2} = 2\int {\dfrac{{{y^2}}}{{1 + {y^2}}}} dy$
Adding and subtracting $1$ in numerator and denominator,
${I_2} = 2\int {\dfrac{{{y^2} + 1 - 1}}{{1 + {y^2}}}} dy$
${I_2} = 2\int {(1 - \dfrac{1}{{1 + {y^2}}}} )dy$
Substituting the value of ${I_2}$in equation $2$,
$I = x{\tan ^{ - 1}}\sqrt x - \dfrac{1}{2}{I_2}$
$I = x{\tan ^{ - 1}}\sqrt x - \dfrac{1}{2} \times 2\int {(1 - \dfrac{1}{{1 + {y^2}}}} )dy$
$I = x{\tan ^{ - 1}}\sqrt x - \int {(1 - \dfrac{1}{{1 + {y^2}}}} )dy$
$I = x{\tan ^{ - 1}}\sqrt x - y + {\tan ^{ - 1}}y - C$ ($\because \int {\dfrac{1}{{1 + {y^2}}}} dy = {\tan ^{ - 1}}y$)
Where $C$is the constant of Integration, We can replace $ - C$with $ + c$.
Substituting the value of $y$,
$I = x{\tan ^{ - 1}}\sqrt x - \sqrt x + {\tan ^{ - 1}}\sqrt x + C$
$I = (x + 1){\tan ^{ - 1}}\sqrt x - \sqrt x + C$

So, the integration of $\int {{{\tan }^{ - 1}}} \sqrt x dx$ is $(x + 1){\tan ^{ - 1}}\sqrt x - \sqrt x + C$.

Note:
Always remember the important formulas of differentiation and integration. If not, then it will make any question more difficult.
Some of the important formula we have used in this question are:
$1)$$\int {\dfrac{1}{{1 + {y^2}}}} dy = {\tan ^{ - 1}}y$
$2)$ Differentiation of ${\tan ^{ - 1}}x = \dfrac{1}{{1 + {x^2}}}$
$3)$$\int {uvdx = u\int {vdx - \int {(u'\int {vdx} } } } )dx$