Question

solve please.........................

$\int \: \bf \: e ^{x \frac{1 + \sin \: x}{1 + \cos \: x} } dx$
​

Answer 1

Question :

Evaluate :

$\begin{gathered}\int \limits_ {0}^{\pi} {e}^{ | \cos x| }[ \: 3 \cos( \frac{1}{2} \cos x) + 2 \sin \: ( \frac{1}{2} \cos x)] \sin x \: dx \\\end{gathered} </p><p>$

Some useful Results used :

$1) \int e {}^{ax} \sin(bx)dx = \frac{e {}^{ax} }{a {}^{2} + b {}^{2} } (a \sin(bx) - b \cos(bx) ) + c1)∫e </p><p>ax$

$2) \int e {}^{ax} \cos(bx)dx = \frac{e {}^{ax} }{a {}^{2} + b {}^{2} } (a \cos(bx) + b \sin(bx) ) + c2)∫e </p><p>$

$let \frac{1}{2} \cos(x)$

$\huge\fcolorbox{black}{aqua}{Solution:-}$

Now differinate with respect to x

$\implies - \sin(x)dx = 2dt</p><p>$

$</p><p>\begin{gathered}\int \limits_ {0}^{\pi} {e}^{ | \cos x| }[ \: 3 \cos( \frac{1}{2} \cos x) + 2 \sin \: ( \frac{1}{2} \cos x)] \sin x \: dx \\\end{gathered}$

$= \bf( \frac{14e}{5} + \frac{14}{5e} )( \sin( \frac{1}{2} ) ) + ( \frac{8e}{5} - \frac{8}{5e})( \cos( \frac{1}{2}$

Answer 2

Question :

Evaluate :

\begin{gathered}\begin{gathered}\int \limits_ {0}^{\pi} {e}^{ | \cos x| }[ \: 3 \cos( \frac{1}{2} \cos x) + 2 \sin \: ( \frac{1}{2} \cos x)] \sin x \: dx \\\end{gathered} < /p > < p > \end{gathered}

0

∫

π

e

∣cosx∣

[3cos(

2

1

cosx)+2sin(

2

1

cosx)]sinxdx

</p><p>

Some useful Results used :

1) \int e {}^{ax} \sin(bx)dx = \frac{e {}^{ax} }{a {}^{2} + b {}^{2} } (a \sin(bx) - b \cos(bx) ) + c1)∫e < /p > < p > ax1)∫e

ax

sin(bx)dx=

a

2

+b

2

e

ax

(asin(bx)−bcos(bx))+c1)∫e</p><p>ax

2) \int e {}^{ax} \cos(bx)dx = \frac{e {}^{ax} }{a {}^{2} + b {}^{2} } (a \cos(bx) + b \sin(bx) ) + c2)∫e < /p > < p >2)∫e

ax

cos(bx)dx=

a

2

+b

2

e

ax

(acos(bx)+bsin(bx))+c2)∫e</p><p>

let \frac{1}{2} \cos(x)let

2

1

cos(x)

\huge\fcolorbox{black}{aqua}{Solution:-}

Solution:-

Now differinate with respect to x

\implies - \sin(x)dx = 2dt < /p > < p >⟹−sin(x)dx=2dt</p><p>

\begin{gathered} < /p > < p > \begin{gathered}\int \limits_ {0}^{\pi} {e}^{ | \cos x| }[ \: 3 \cos( \frac{1}{2} \cos x) + 2 \sin \: ( \frac{1}{2} \cos x)] \sin x \: dx \\\end{gathered} \end{gathered}

</p><p>

0

∫

π

e

∣cosx∣

[3cos(

2

1

cosx)+2sin(

2

1

cosx)]sinxdx

= \bf( \frac{14e}{5} + \frac{14}{5e} )( \sin( \frac{1}{2} ) ) + ( \frac{8e}{5} - \frac{8}{5e})( \cos( \frac{1}{2}=(

5

14e

+

5e

14

)(sin(

2

1

))+(

5

8e

−

5e

8

)(cos(

2

1