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Question

Show that dxsinx(5+4cosx)=118log(1cosx)12log(1+cosx)+49log(5+4cosx).

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Solution

Let I=1sinx(5+4cosx)dx
Multiplying numerator and denominator by sinx, we get
I=sinx(4cosx+5)(cos2x1)dx
Put t=cosx
dt=sinxdx

So, I=1(4t+5)(t21)dt=1(4t+5)(t+1)(t1)

Resolving 1(4t+5)(t+1)(t1) into partial fractions

1(4t+5)(t+1)(t1)=A4t+5+Bt+1+Ct1 ....(1)

1=A(t21)+B(t1)(4t+5)+C(4t+5)(t+1) ....(2)

Put t=1 in eqn (2)
C=118
Put t=1 ineqn (2)
B=12
Put t=54
A=169

Put these values in eqn (1)
1(4t+5)(t+1)(t1)=169(4t+5)+12(t+1)+118(t1)

I=1(4t+5)(t21)dt=169dt(4t+5)12dt(t+1)+118dt(t1)
=12log(t+1)+169×14log(4t+5)+118log(t1)+C
=12log(cosx+1)+49log(4cosx+5)+118log(cosx1)+C

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