integration.cotx ×dx÷1+sin square x=
Answers
∫cot(x)1+sin2(x)dx
cot(x)=cos(x)sin(x)
So lets put that into the integral:
=∫cos(x)sin(x)1+sin2(x)dx
=∫cos(x)sin(x)(1+sin2(x))dx
Lets multiply everything by: sin(x)sin(x)
=∫cos(x)sin(x)(1+sin2(x))⋅sin(x)sin(x)dx
=∫cos(x)sin(x)sin2(x)(1+sin2(x))dx
Suppose u=sin2(x)+1, thus dudx=2sin(x)cos(x)
du=2sin(x)cos(x)dx
dx=du2sin(x)cos(x)
=∫cos(x)sin(x)sin2(x)(1+sin2(x))dx
=∫cos(x)sin(x)(sin2(x)+1−1)(1+sin2(x))dx
=∫12(u−1)(u)du
=12∫1u(u−1)du
Lets now use partial fraction decomposition to split the fraction into two fractions which can be integrated easily:
1u(u−1)=Au+Bu−1
1u(u−1)=Au⋅u−1u−1+Bu−1⋅uu
1u(u−1)=Au−Au(u−1)+Buu(u−1)
1u(u−1)=Au−A+Buu(u−1)
1=Au−A+Bu
1=u(A+B)−A
A+B=0∧A=−1
−1+B=0
B=1
Now we know:
1u(u−1)=−1u+1u−1
Lets put it back into the integral:
=12∫1u(u−1)du
=12∫−1u+1u−1du
=12∫1u−1−1udu
=12∫1u−1du−12∫1udu
=12ln∣∣u−1∣∣−12ln∣∣u∣∣+C
Lets put back u=sin2(x)+1
=12ln∣∣sin2(x)+1−1∣∣−12ln∣∣sin2(x)+1∣∣+C
=12ln∣∣sin2(x)∣∣−12ln∣∣sin2(x)+1∣∣+C
=2⋅12ln∣∣sin(x)∣∣−12ln∣∣sin2(x+1∣∣+C
=ln∣∣sin(x)∣∣−12ln∣∣sin2(x)+1∣∣+C
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