Math, asked by Anonymous, 1 day ago

Integrate by Parts :-

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

Answers

Answered by ValeryLegasov
5

Answer:

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

Attachments:
Answered by mathdude500
8

\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}

Similar questions