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secondary 4 | A Maths
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Teck wee
Teck Wee

secondary 4 chevron_right A Maths chevron_right Singapore

Take i as an unknown.

Date Posted: 1 month ago
Views: 20
J
J
1 month ago
Trick is to rearrange the terms such that it directly is in the form of the derivative contained using quotient rule. Then it can be directly integrated.



sinx e^(cosx) - (sinx + cosx)e^(sinx + cosx)
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
e^(2sinx) - 2e^(sinx) + 1

=

sinx e^(cosx) - (sinx + cosx) e^(sinx) e^(cosx)
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
(e^(sinx))² - 2e^(sinx) + 1

=

sinxe^(cosx) -sinx e^(sinx)e^(cosx) -cosx e^(sinx)e^(cosx)
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
(e^(sinx)) - 1)²

=

sinx e^(cosx) (1 - e^(sinx) ) - (cosx e^(sinx)) e^(cosx)
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
(e^(sinx)) - 1)²


=

-sinx e^(cosx) (e^(sinx) - 1) - e^(cosx) (cosx e^(sinx))
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
(e^(sinx)) - 1)²



Notice that this is in the form

du/dx (v) - (u) dv/dx
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾


= d/dx (u/v)


whereby :

v = e^(sinx) - 1
dv/dx = cosx e^(sinx)

u = e^(cosx)
du/dx = - sinx e^(cosx)



So, u/v = e^(cosx)/(e^(sinx) - 1)


And therefore integration of the expression gives us

e^(cosx)/(e^(sinx) - 1) + c , c is a constant

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Nalin Sharma
Nalin Sharma's answer
35 answers (Tutor Details)
1st
An ansatz solution, don't know if this acceptable in the O Level, but gives you the answer for sure.