The question is, find the integral to the

zdebe5l8

zdebe5l8

Answered question

2022-04-09

The question is, find the integral to the function:
sin3θsin3θcos3θ

Answer & Explanation

Gonarsu2dw8

Gonarsu2dw8

Beginner2022-04-10Added 19 answers

I=sin3(x)sin3(x)cos3(x)dx=tan3(x)tan3(x)1dx
Let t=tan(x). Then dt=sec2(x)dx. Since tan(x)=t, we have sec2(x)=1+tan2(x)=1+t2 and hence dx=dtsec2(x)=dt1+t2
Now the integral becomes
I=t3(t31)(1+t2)dt
Now resort to the good old partial fractions to get the integral.
folytonr4qx

folytonr4qx

Beginner2022-04-11Added 14 answers

We take advantage of the symmetry, indeed expand on it. Let
I=sin3θdθsin3θcos3θ and J=cos3θdθsin3θcos3θ
Note that
sin3θsin3θcos3θ=1+cos3θsin3θcos3θ
and therefore
IJ=θ
If we can find I+J we will be finished. So we want to find
sin3θ+cos3θsin3θcos3θ,dθ=(sinθ+cosθ)(sin2θ+cos2θsinθcosθ)(sinθcosθ)(sin2θ+cos2θ+sinθcosθ),dθ
Let u=sinθcosθ Then du=(cosθ+sinθ)dθ. Also u2=12sinθcosθ. From this we find that sin2θ+cos2θsinθcosθ=1+u22 and sin2θ+cos2θ+sinθcosθ=3u22. Thus
I+J=1+u2u(3u2),du
We do a partial partial fraction decomposition:
1+u2u(3u2)=13(1u+4u3u2)
Integrate: I+J=(13)ln(|u|(3u2)2)

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