i am at a point where i have to do 10 trigonometric substitutions.
wtf, are they expecting me to do?
are they ******* insane?
wtf, are they expecting me to do?
are they ******* insane?
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"are they ******* insane?" Probably. Stupid is more like it.
It is a sign of a poor teacher to give long, meaningless calculations.
The point can be brought home by integrating sin^3(x), much shorter and carrying the same message.
It is a sign of a poor teacher to give long, meaningless calculations.
The point can be brought home by integrating sin^3(x), much shorter and carrying the same message.
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You can do it a little faster by playing with complex numbers.
sin^6(x)
= [ exp(ix) - exp(-ix) ]^6 / (2i)^6
= [exp(6ix) - 6exp(4ix) + 15exp(2ix) - 20 + 15exp(-2ix) - 6exp(-4ix) + exp(-6ix)] / (-64)
= [ 2 cos(6x) - 12 cos(4x) + 30 cos(2x) - 20 ] / (-64)
= [ cos(6x) - 6 cos(4x) + 15 cos(2x) - 10 ] / (-32)
The integral is
[ sin(6x)/6 - (3/2) sin(4x) + (15/2) sin(2x) - 10x ] / (-32)
sin^6(x)
= [ exp(ix) - exp(-ix) ]^6 / (2i)^6
= [exp(6ix) - 6exp(4ix) + 15exp(2ix) - 20 + 15exp(-2ix) - 6exp(-4ix) + exp(-6ix)] / (-64)
= [ 2 cos(6x) - 12 cos(4x) + 30 cos(2x) - 20 ] / (-64)
= [ cos(6x) - 6 cos(4x) + 15 cos(2x) - 10 ] / (-32)
The integral is
[ sin(6x)/6 - (3/2) sin(4x) + (15/2) sin(2x) - 10x ] / (-32)
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you could do an integration by parts [ sin x ]^6 dx = [ sinx ]^5 { - cos x } / 6 +[5/6] { [sin x]^4 dx }
integrating the last by parts will take you to integration of sin² x { = [1-cos 2x]/2 dx }
u = sin^5 x , dv = sin x dx to start...try it
integrating the last by parts will take you to integration of sin² x { = [1-cos 2x]/2 dx }
u = sin^5 x , dv = sin x dx to start...try it
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no it's not you and yeah it's kind of crazy but good practice i guess. There are tables you could use but they want you to grind it out when you're first learning it. I know it sucks...
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it takes a while. don't worry, you're probably doing it right. it'll take like a full page or more.