Math, asked by chandavedic, 22 days ago

Solve the following integration

 \int \:  \frac{sinx - xcosx}{ {x}^{2} + xsinx} dx

Answers

Answered by mathdude500
42

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

Given integral is

\rm \: \displaystyle\int\rm  \frac{sinx - xcosx}{ {x}^{2} + xsinx}dx \\

can be rewritten as

\rm \: =  \:  \displaystyle\int\rm  \frac{sinx - xcosx}{x(x + sinx)}dx \\

\rm \: =  \:  \displaystyle\int\rm  \frac{sinx + x - x - xcosx}{x(x + sinx)}dx \\

\rm \: =  \:  \displaystyle\int\rm  \frac{(sinx + x) - x(1 + cosx)}{x(x + sinx)}dx \\

\rm \:  =  \: \displaystyle\int\rm  \frac{x + sinx}{x(x + sinx)}dx \:  -  \: \displaystyle\int\rm  \frac{x(1 + cosx)}{x(x + sinx)}dx \\

\rm \:  =  \: \displaystyle\int\rm  \frac{1}{x} \: dx \:  -  \: \displaystyle\int\rm  \frac{1 + cosx}{x + sinx} \: dx \\

Now, in second integral

\boxed{ \rm{ \: \frac{d}{dx}(x + sinx) = 1 + cosx \: }} \\

and we know

\boxed{ \rm{ \:\displaystyle\int\rm  \frac{f'(x)}{f(x)} \: dx \:  =  \: log |f(x)| + c \:  \: }} \\

So, using this, we get

\rm \:  =  \: log |x|  -  log |x + sinx| + c \\

Hence,

\color{green}\rm\implies \:\rm \: \displaystyle\int\rm  \frac{sinx - xcosx}{ {x}^{2} + xsinx}dx \\ \\ \color{green} \rm \:  =  \: log |x|  - log |x + sinx|  + c

or

\color{blue}\rm\implies \:\rm \: \displaystyle\int\rm  \frac{sinx - xcosx}{ {x}^{2} + xsinx}dx \\ \\ \color{blue} \rm \:  =  \: log \bigg| \frac{x}{x + sinx}\bigg |  + 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}

Answered by Anonymous
72

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

Given Integral

  \displaystyle\rm{ \int \: \frac{sinx - xcosx}{ {x}^{2} + xsinx} dx}

Now,

{ \implies\displaystyle \rm{ \int \:  \frac{ (\sin  x + x - x - x \cos x)}{x (x + \sin x)} }}

{ \implies \displaystyle \rm{ \int \:  \frac{(x +  \sin x)}{x(x +  \sin x)} -  \frac{x(1 +  \cos x)}{x(x +  \sin x)} dx}}

{ \implies \displaystyle \rm{ \int  \frac{1}{x}dx -  \int \:  \frac{1 +  \cos x}{x +  \sin x}  dx}}

{ \implies \displaystyle \rm{ =  \log |x| -   \int \:  \frac{dp}{p}  }}

{ \boxed{ \implies \displaystyle \rm{ =  \log |x| -  \log |p| + c  }}}

{ \implies \displaystyle \rm{ \log |x| -  \log  |x +  \sin x | + c   }}

{ \implies{ \boxed {\displaystyle {\rm{ \log \:   | \frac{x}{x +  \sin x} |  + c}}}}}

_________________________

Additional Information

\displaystyle\sf\;\dfrac{d}{dx}(sin\;x)=cosx \\\\  \sf \;\dfrac{d}{dx}(cos\;x) = -sinx \\\\ \sf\; \dfrac{d}{dx}(tan\;x) = sec^{2}x \\\\\sf \; \dfrac{d}{dx}(cot\;x) = -csc^{2}x \\\\ \sf \; \dfrac{d}{dx}(sec\;x) = secx \cdot tanx \\\\ \sf \; \dfrac{d}{dx}(csc\;x) = -cscx \cdot cotx \\\\ \sf\; \dfrac{d}{dx}(sinh\;x)=coshx \\\\ \sf\; \dfrac{d}{dx}(cosh\;x)= sinhx \\\\\sf \;\dfrac{d}{dx}(tanh\;x)=sech^{2}h \\\\\sf\;\dfrac{d}{dx}(coth\;x)=-csch^{2}x \\\\ \sf\;\dfrac{d}{dx}(sech\;x) =-sechx \cdot tanhx \\\\ \sf\;\dfrac{d}{dx}(csch\;x) = -cschx \cdot cothx

Similar questions