Ok here it is,(adsbygoogle = window.adsbygoogle || []).push({});

Show that the integral of

sin(θ)-cos(θ)dθ

sin(θ)+cos(θ)

= -(1/2)ln(2sin(2θ)+2)+C

Now what I did was let u=sin(θ)+cos(θ) so that du=cos(θ)-sin(θ)dθ or that du=-(sin(θ)-cos(θ))dθ

Now I got the integral of

(sin(θ)-cos(θ))*du

u*-(sin(θ)-cos(θ))

And then both of the sin(θ)-cos(θ) cancels out and I was left with the integral of

-du

u

which is -ln|u|+C , plugging back what u was into that I got

-ln|sin(θ)+cos(θ)|+C

which I cannot figure out to be equal to what he said it should be.

The only thing I can think of is the C's being different and that causing them to be equal. Any ideas?

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# Math problem

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