; 01-31-2019 05:04 PM
6) 1/2e^4
8) o
10) 0
\[\begin{array}{l}
Q8\\
\int\limits_{ - \infty }^\infty {\frac{x}{{\sqrt {{x^2} + 2} }}} \,\,dx\\
Solution:\int\limits_{ - \infty }^\infty {\frac{x}{{\sqrt {{x^2} + 2} }}} \,\,dx = \int\limits_{ - \infty }^0 {\frac{x}{{\sqrt {{x^2} + 2} }}} \,dx + \int\limits_0^\infty {\frac{x}{{\sqrt {{x^2} + 2} }}} \,\,dx\\
1)\,\,\int\limits_{ - \infty }^0 {\frac{x}{{\sqrt {{x^2} + 2} }}} \,\,dx = \mathop {\lim }\limits_{b \to - \infty } \int\limits_b^0 {\frac{x}{{\sqrt {{x^2} + 2} }}} \,\,dx = \mathop {\lim }\limits_{b \to - \infty } \left. {\sqrt {{x^2} + 2} } \right]\begin{array}{*{20}{c}}
0\\
b
\end{array}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{b \to - \infty } \left. {\left( {\sqrt 2 - \sqrt {{b^2} + 2} } \right)} \right] = \sqrt 2 - \infty = - \infty \,\,\\
2)In\,same\,way\,\,\,\,\int\limits_0^\infty {\frac{x}{{\sqrt {{x^2} + 2} }}} \,\,dx = \mathop {\lim }\limits_{a \to \infty } \int\limits_0^a {\frac{x}{{\sqrt {{x^2} + 2} }}} \,\,dx = \mathop {\lim }\limits_{a \to \infty } \left. {\sqrt {{x^2} + 2} } \right]\begin{array}{*{20}{c}}
a\\
0
\end{array}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{a \to \infty } \left. {\left( {\sqrt {{a^2} + 2} - \sqrt 2 } \right)} \right] = \infty - \sqrt 2 = \infty \,\,\\
Hence\,\,\int\limits_{ - \infty }^\infty {\frac{x}{{\sqrt {{x^2} + 2} }}} \,\,dx\,\,\,is\,diverges
\end{array}\]
; 02-03-2019 06:42 PM
; 02-03-2019 06:31 PM
; 02-03-2019 06:35 PM
\[\begin{array}{l}
Evaluate:\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\,\,\int {{{\tan }^{ - 1}}\left( {\frac{{\sin x}}{{1 + \cos x}}} \right)} \,dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,K1
\end{array}\]
; 02-20-2019 01:13 AM
Tan (x/2)=sin x/(1+cos x) implies the integral=integral of x/2=1/4x^2 +c
Q1 kamil.jpg
; 02-24-2019 12:52 AM
(Tan (pi/4+x/2)=cos x/(1+sin x)
=then the integral= integral of pi/4+x/2
Pi/4 x+x^2/4+c
: 1 (0 1 )
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