## Question

secondary 4 | A Maths

Anyone can contribute an answer, even non-tutors.

##### Teck Wee

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Take i as an unknown.

Date Posted: 11 months ago
Views: 83
J
11 months 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