Math, asked by madhav5245, 4 days ago

Solve the following

 \int \:  \frac{(x + 1) {e}^{x} }{ {sin}^{2} (x {e}^{x} )}  \:d x

Answers

Answered by xxblackqueenxx37
62

Important topic :-

The following question is related to topics trigonometry for this topic it is very necessary to learn Pythagoras theorem and the table value of trigonometry ratios of angles 0° , 30°, 45°, 60°, 90° {which is shared in following image}

Given :-

  •  \sf \: \int \: \frac{(x + 1) {e}^{x} }{ {sin}^{2} (x {e}^{x} )} \:d x \\

Solution :-

 \sf \: =  let \:  {xe}^{x}  = t

 \sf \:  = ( {e}^{x}  +  {xe}^{x} )dx = dt

 \sf \:  =  {e}^{x} (1 + x)dx = dt

 \sf \:  = \int \: \frac{dt}{ {sin}^{2} t} \: =  \int \: cose \:  {c}^{2}  \: tdt \\

 \sf \:  =  - cot(t) + c

 \sf \:  =  - cot(x {e}^{x} ) + c

 \sf \:  = \int \: \frac{(x + 1) {e}^{x} }{ {sin}^{2} (x {e}^{x} )} \:d x =  -  \cot(x {e}^{x} )  + c \\

  \sf \fbox \red{Answer  }  = \:  \: \sf  \int \: \frac{(x + 1) {e}^{x} }{ {sin}^{2} (x {e}^{x} )} \:d x =  \sf  \red{  -  \cot(x {e}^{x} )  + c} \\

Note :-

scroll the page to (→) direction to see the full Answer

Addictonal information :-

There are 6 trigonometric functions which are:

  • Sine function
  • Cosine function
  • Tan function
  • Sec function
  • Cot function
  • Cosec function

warning :-

If you are writing this in you class book please don't write Important topic, Note and addictonal information.

Attachments:
Answered by mathdude500
37

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

Given integral is

\rm \: \displaystyle\int\rm  \frac{(x + 1){e}^{x}}{sin^{2} (x{e}^{x})}  \: dx \\

To evaluate this integral, we use method of Substitution.

So, Substitute

 \red{\rm \: x{e}^{x} = y \: } \\

On differentiating both sides w. r. t. x, we get

 \red{\rm \: x{e}^{x} + {e}^{x} \times 1 =  \dfrac{dy}{dx} \: } \\

 \red{\rm \: x{e}^{x} + {e}^{x} =  \dfrac{dy}{dx} \: } \\

 \red{\rm \: (x +1) {e}^{x} =  \dfrac{dy}{dx} \: } \\

 \red{\rm \: (x +1) {e}^{x}dx =  dy \: } \\

So, on substituting these values in above integral, we get

\rm \:  =  \: \displaystyle\int\rm  \frac{dy}{ {sin}^{2} y}  \\

\rm \:  =  \: \displaystyle\int\rm  {cosec}^{2}y \: dy \\

\rm \:  =  \:  -  \: coty \:  +  \: c \\

\rm \:  =  \:  -  \: cot(x{e}^{x}) \:  +  \: c \\

Hence,

\rm\implies \:\boxed{ \rm{ \:\displaystyle\int\rm  \frac{(x + 1){e}^{x}}{ {sin}^{2}(x{e}^{x}) } dx  =  \:  -  \: cot(x{e}^{x}) \:  +  \: c  \: }}\\

\rule{190pt}{2pt}

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}

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