The question is, find the integral to the function:sin3θsin3θ−cos3θ

Question

Gonarsu2dw8 · Accepted Answer

I=∫sin3(x)sin3(x)−cos3(x)dx=∫tan3(x)tan3(x)−1dxLet 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+t2Now the integral becomesI=∫t3(t3−1)(1+t2)dtNow resort to the good old partial fractions to get the integral.

folytonr4qx · Accepted Answer

We take advantage of the symmetry, indeed expand on it. Let

I = \int \frac{\sin^{3} θ d θ}{\sin^{3} θ - \cos^{3} θ}

and

J = \int \frac{\cos^{3} θ d θ}{\sin^{3} θ - \cos^{3} θ}

Note that

\frac{\sin^{3} θ}{\sin^{3} θ - \cos^{3} θ} = 1 + \frac{\cos^{3} θ}{\sin^{3} θ - \cos^{3} θ}

and therefore

I - J = θ

If we can find

I + J

we will be finished. So we want to find

\int \frac{\sin^{3} θ + \cos^{3} θ}{\sin^{3} θ - \cos^{3} θ}, d θ = \int \frac{(\sin θ + \cos θ) (\sin^{2} θ + \cos^{2} θ - \sin θ \cos θ)}{(\sin θ - \cos θ) (\sin^{2} θ + \cos^{2} θ + \sin θ \cos θ)}, d θ

Let

u = \sin θ - \cos θ

Then

d u = (\cos θ + \sin θ) d θ

. Also

u^{2} = 1 - 2 \sin θ \cos θ

. From this we find that

\sin^{2} θ + \cos^{2} θ - \sin θ \cos θ = \frac{1 + u^{2}}{2}

and

\sin^{2} θ + \cos^{2} θ + \sin θ \cos θ = \frac{3 - u^{2}}{2}

. Thus

I + J = \int \frac{1 + u^{2}}{u (3 - u^{2})}, d u

We do a partial partial fraction decomposition:

\frac{1 + u^{2}}{u (3 - u^{2})} = \frac{1}{3} (\frac{1}{u} + \frac{4 u}{3 - u^{2}})

Integrate:

I + J = (\frac{1}{3}) \ln (\frac{| u |}{{(3 - u^{2})}^{2}})

The question is, find the integral to the

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New Questions in Calculus 1