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## Question

secondary 4 | A Maths

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Take i as an unknown.

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

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

v²

= 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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