Chemistry, asked by moolyadikshita, 1 month ago

the electron wave in an atom is of type​

Answers

Answered by bhagyashreehappy123
0

Answer:

\textbf{Given:}Given:

\mathsf{\int\,sec^nu\;du}∫sec

n

udu

\textbf{To find:}To find:

\textsf{Reduction formula for}\;\mathsf{\int\,sec^nu\;du}Reduction formula for∫sec

n

udu

\textbf{Solution:}Solution:

\textsf{Consider,}Consider,

\mathsf{\int\,sec^nu\;du}∫sec

n

udu

\textsf{We apply Integration by parts formula}We apply Integration by parts formula

\boxed{\mathsf{\int\,m\,dn=mn-\int\,n\;dm}}

∫mdn=mn−∫ndm

\mathsf{=\int\,sec^{n-2}u\,(sec^2u\,du)}=∫sec

n−2

u(sec

2

udu)

\mathsf{Take,}Take,

\mathsf{m=sec^{n-2}u\;\implies\;dm=(n-2)\,sec^{n-3}u\;du}m=sec

n−2

u⟹dm=(n−2)sec

n−3

udu

\mathsf{dn=sec^u\,du\;\implies\;\int\,dn=\int\,sec^2u\,du\;\implies\;n=tanu}dn=sec

u

du⟹∫dn=∫sec

2

udu⟹n=tanu

\mathsf{Now,}Now,

\mathsf{\int\,sec^nu\;du=sec^{n-2}u\,tanu-\int\,tanu\,(n-2)sec^{n-3}u\,(secu\,tanu)du}∫sec

n

udu=sec

n−2

utanu−∫tanu(n−2)sec

n−3

u(secutanu)du

\mathsf{\int\,sec^nu\;du=sec^{n-2}u\,tanu-\int\,tanu\,(n-2)sec^{n-3}u(secu\,tanu)du}∫sec

n

udu=sec

n−2

utanu−∫tanu(n−2)sec

n−3

u(secutanu)du

\mathsf{\int\,sec^nu\;du=sec^{n-2}u\,tanu-(n-2)\int\,tan^2u\,sec^{n-2}u\,du}∫sec

n

udu=sec

n−2

utanu−(n−2)∫tan

2

usec

n−2

udu

\mathsf{\int\,sec^nu\;du=sec^{n-2}u\,tanu-(n-2)\int(sec^2u-1)\,sec^{n-2}u\,du}∫sec

n

udu=sec

n−2

utanu−(n−2)∫(sec

2

u−1)sec

n−2

udu

\mathsf{\int\,sec^nu\;du=sec^{n-2}u\,tanu-(n-2)\int(sec^nu-sec^{n-2}u)du}∫sec

n

udu=sec

n−2

utanu−(n−2)∫(sec

n

u−sec

n−2

u)du

\mathsf{\int\,sec^nu\;du=sec^{n-2}u\,tanu-(n-2)\int\,sec^nu\,du+(n-2)\int\,sec^{n-2}udu}∫sec

n

udu=sec

n−2

utanu−(n−2)∫sec

n

udu+(n−2)∫sec

n−2

udu

\mathsf{\int\,sec^nu\;du+(n-2)\int\,sec^nu\,du=sec^{n-2}u\,tanu+(n-2)\int\,sec^{n-2}udu}∫sec

n

udu+(n−2)∫sec

n

udu=sec

n−2

utanu+(n−2)∫sec

n−2

udu

\mathsf{(n-1)\int\,sec^nu\,du=sec^{n-2}u\,tanu+(n-2)\int\,sec^{n-2}udu}(n−1)∫sec

n

udu=sec

n−2

utanu+(n−2)∫sec

n−2

udu

\implies\boxed{\mathsf{\int\,sec^nu\,du=\dfrac{sec^{n-2}u\,tanu}{n-1}+\dfrac{n-2}{n-1}\int\,sec^{n-2}udu}}⟹

∫sec

n

udu=

n−1

sec

n−2

utanu

+

n−1

n−2

∫sec

n−2

udu

\textbf{Find more:}Find more:

Answered by Anonymous
0

Answer:

Here's the ans hope it's helpful for you

Explanation:

Along with all other quantum objects, an electron is partly a wave and partly a particle. To be more accurate, an electron is neither literally a traditional wave nor a traditional particle, but is instead a quantized fluctuating probability wavefunction.

#its Sayan

Similar questions