Question

integration of e^x (cosx-sinx)cosec^2 x dx

Answer 1

$\textbf{To find:}$

$\int\,e^x(cosx-sinx)cosec^2x\,dx$

$\text{Consider,}$

$\int\,e^x(cosx-sinx)cosec^2x\,dx$

$=\int\,e^x(cosx\,cosec^2x-sinx\,cosec^2x)\,dx$

$=\int\,e^x(\frac{1}{sinx}\,\frac{cosx}{sinx}-sinx\,\frac{1}{sin^2x})\,dx$

$=\int\,e^x(cosecx\,cotx-cosecx)\,dx$

$=\int\,e^x(-cosecx+cosecx\,cotx)\,dx$

$\boxed{\begin{minipage}{4cm}\textbf{Take } \bf\,f(x)=-cosecx\\\\\bf\,f'(x)=cosecx\,cotx \end{minipage}}$

$=\int\,e^x[(f(x)+f'(x)]\,dx$

$\text{Using}$

$\boxed{\bf\,\int\,e^x[(f(x)+f'(x)]\,dx=e^x\,f(x)+c}$

$=e^x\,f(x)+c$

$=e^x\,(-cosecx)+c$

$=-e^x\,cosecx+c$

$\therefore\bf\int\,e^x(cosx-sinx)cosec^2x\,dx=-e^x\,cosecx+c$

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Integrate of ex sec x(1+tanx) dx

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Answer 2

Answer:

\textbf{To find:}To find:

\int\,e^x(cosx-sinx)cosec^2x\,dx∫e

x

(cosx−sinx)cosec

2

xdx

\text{Consider,}Consider,

\int\,e^x(cosx-sinx)cosec^2x\,dx∫e

x

(cosx−sinx)cosec

2

xdx

=\int\,e^x(cosx\,cosec^2x-sinx\,cosec^2x)\,dx=∫e

x

(cosxcosec

2

x−sinxcosec

2

x)dx

=\int\,e^x(\frac{1}{sinx}\,\frac{cosx}{sinx}-sinx\,\frac{1}{sin^2x})\,dx=∫e

x

(

sinx

1

sinx

cosx

−sinx

sin

2

x

1

)dx

=\int\,e^x(cosecx\,cotx-cosecx)\,dx=∫e

x

(cosecxcotx−cosecx)dx

=\int\,e^x(-cosecx+cosecx\,cotx)\,dx=∫e

x

(−cosecx+cosecxcotx)dx

\begin{gathered}\boxed{\begin{minipage}{4cm}\textbf{Take }$\bf\,f(x)=-cosecx\\\\\bf\,f'(x)=cosecx\,cotx$\end{minipage}}\end{gathered}

=\int\,e^x[(f(x)+f'(x)]\,dx=∫e

x

[(f(x)+f

′

(x)]dx

\text{Using}Using

\boxed{\bf\,\int\,e^x[(f(x)+f'(x)]\,dx=e^x\,f(x)+c}

∫e

x

[(f(x)+f

′

(x)]dx=e

x

f(x)+c

=e^x\,f(x)+c=e

x

f(x)+c

=e^x\,(-cosecx)+c=e

x

(−cosecx)+c

=-e^x\,cosecx+c=−e

x

cosecx+c

\therefore\bf\int\,e^x(cosx-sinx)cosec^2x\,dx=-e^x\,cosecx+c∴∫e

x

(cosx−sinx)cosec

2

xdx=−e

x

cosecx+c

Find more:

Integrate of ex sec x(1+tanx) dx

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