Question

Integrate by Parts :-

$\quad \qquad { \bigstar { \underline { \boxed { \bf { \red { \int {e}^{ {Cos}^{- 1} x } }}}}}}{\bigstar }$ ​

Answer 1

Answer:

i solved it using integration by parts check the images for solution

Answer 2

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm {e}^{ {cos}^{ - 1}x } \: dx$

Let assume that

$\rm :\longmapsto\:I = \displaystyle\int\rm {e}^{ {cos}^{ - 1}x } \: dx$

Now to evaluate this integral, we use Method of Substitution.

So, Substitute,

$\red{\rm :\longmapsto\: {cos}^{ - 1}x = y}$

$\red{\rm :\longmapsto\:x = cosy}$

$\red{\rm :\longmapsto\:dx = - \: siny \: dy}$

On substituting all these values, we get

$\rm :\longmapsto\:I = \displaystyle\int\rm {e}^{y} \: ( - siny) \: dy$

$\rm :\longmapsto\:I \: = - \: \displaystyle\int\rm {e}^{y} \: siny \: dy$

So, using integration by parts, we get

$\rm \:I = \: - siny\displaystyle\int\rm {e}^{y}dy + \displaystyle\int\rm \bigg[\dfrac{d}{dy}siny\displaystyle\int\rm {e}^{y}dy \bigg]dy$

$\rm \: I = \: - siny \: {e}^{y}+ \displaystyle\int\rm cosy \: {e}^{y} dy$

Again, using integration by parts, we get

$\rm \: I = \: - siny \: {e}^{y}+cosy \displaystyle\int\rm {e}^{y} dy - \displaystyle\int\rm \bigg[\dfrac{d}{dy} cosy\displaystyle\int\rm {e}^{y}dy\bigg]dy$

$\rm \: I = \: - siny \: {e}^{y}+cosy \: {e}^{y} + \displaystyle\int\rm siny \: {e}^{y} + c$

$\rm :\longmapsto\: I = \: - siny \: {e}^{y}+cosy \: {e}^{y} - I + c$

$\rm :\longmapsto\:2 I = \: - siny \: {e}^{y}+cosy \: {e}^{y} + c$

$\rm :\longmapsto\:I = \: \dfrac{1}{2} \bigg( - siny \: {e}^{y}+cosy \: {e}^{y} + c\bigg)$

$\rm :\longmapsto\:I = \: \dfrac{{e}^{y}}{2} \bigg( - siny \:+cosy \:\bigg) + d$

$\rm :\longmapsto\:I = \: \dfrac{{e}^{ {cos}^{ - 1} x}}{2} \bigg( - \sqrt{1 - {cos}^{2} y} \:+x \:\bigg) + d$

$\rm :\longmapsto\:I = \: \dfrac{{e}^{ {cos}^{ - 1} x}}{2} \bigg( - \sqrt{1 - {x}^{2} } \:+x \:\bigg) + d$

where d = c/2.

Hence,

$\boxed{ \tt{ \: \displaystyle\int\rm {e}^{ {cos}^{ - 1} x} = \: \dfrac{{e}^{ {cos}^{ - 1} x}}{2} \bigg(x - \sqrt{1 - {x}^{2} } \:\bigg) + d \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Formula used :-

$\boxed{ \tt{ \: \displaystyle\int\rm {e}^{x} \: dx \: = \: {e}^{x} + c \: }}$

$\boxed{ \tt{ \: \dfrac{d}{dx}sinx = cosx \: }}$

$\boxed{ \tt{ \: \dfrac{d}{dx}cosx \: = \: - \: sinx \: }}$

$\boxed{ \tt{ \: \displaystyle\int\rm uvdx = u\displaystyle\int\rm vdx - \displaystyle\int\rm \bigg[\dfrac{d}{dx} u\displaystyle\int\rm vdx\bigg]dx \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$