8

The integral is:

$$\int \frac{(\sin^2(x)\cdot \cos(x))}{\sin(x)+\cos(x)}dx$$

I used weierstraß substitution

$$t:=\tan(\frac{x}{2})$$

$$\sin(x)=\frac{2t}{1+t^2}$$ $$\cos(x)=\frac{1-t^2}{1+t^2}$$ $$dx=\frac{2}{1+t^2}dt$$

Got this:

$$\int \frac{8t^4-8t^2}{t^8−2t^7+2t^6−6t^5−6t^3−2t^2−2t−1}dt$$

and with partial fraction expansion the final answer is:

$$\frac{1}{4}[\ln(\sin(x)+\cos(x))-\cos(x)*(\sin(x)+\cos(x))]+C$$

It is a long way and I am very convinced, there is a shorter way, maybe you know one? Thanks

Yuriy S
  • 31,474
marius
  • 145
  • 6

3 Answers3

10

If you multiply numerator and denominator by $\cos x-\sin x$, the numerator can be rewritten as $$ \sin x\cos x(\sin x\cos x-\sin^2x) $$ Now use $\sin x\cos x=\frac{1}{2}\sin 2x$ and $$ \sin^2x=\frac{1-\cos2x}{2} $$ so finally we get $$ \frac{1}{4}\sin2x(\sin2x-1+\cos2x)= \frac{1}{4}(1-\cos^22x-\sin2x+\sin2x\cos2x) $$ and the integral becomes $$ \frac{1}{4}\int\left( \frac{1}{\cos2x}-\frac{\sin2x}{\cos2x}-\cos2x+\sin2x \right)\,dx $$ that should pose little problems.

egreg
  • 238,574
5

\begin{align} \int\frac{\sin^2 x\cos x}{\sin x+\cos x}\,\mathrm dx &= \int\frac{\sin^2 x\sin\left(\frac\pi2-x\right)}{\sin x+\sin\left(\frac\pi2-x\right)}\,\mathrm dx \\ &= \int\frac{\sin^2\left(\frac\pi4+u\right)\sin\left(\frac\pi4-u\right)}{\sin \left(\frac\pi4+u\right)+\sin\left(\frac\pi4-u\right)}\,\mathrm d\left(\frac\pi4+u\right) \\ &= \int\frac{\left(\sin\frac\pi4\cos u+\cos\frac\pi4\sin u\right)^2\left(\sin\frac\pi4\cos u-\cos\frac\pi4\sin u\right)}{\left(\sin\frac\pi4\cos u+\cos\frac\pi4\sin u\right)+\left(\sin\frac\pi4\cos u-\cos\frac\pi4\sin u\right)}\,\mathrm d\left(\frac\pi4+u\right) \\ &= \frac14\int\frac{\left(\cos u+\sin u\right)^2\left(\cos u-\sin u\right)}{\cos u}\,\mathrm d\left(\frac\pi4+u\right) \\ &= \frac14\int\frac{\left(\cos u+\sin u\right)\left(\cos^2 u-\sin^2 u\right)}{\cos u}\,\mathrm d\left(\frac\pi4+u\right) \\ &= \frac14\int\left(\cos2u+\sin2u-\tan u\right)\,\mathrm d\left(\frac\pi4+u\right) \\ &= \frac14\left(\frac12\sin\left(2x-\frac\pi2\right)-\frac12\cos\left(2x-\frac\pi2\right)+\log\cos\left(x-\frac\pi4\right)\right)+\textsf{const.} \\ &= \frac14\left(\log\cos\left(x-\frac\pi4\right)-\frac12\left(\sin2x+\cos2x\right)\right)+\textsf{const.} \\ &= \frac14\left(\log\cos\left(x-\frac\pi4\right)-\frac1{\sqrt2}\cos\left(2x-\frac\pi4\right)\right)+\textsf{const.} \end{align}

joriki
  • 238,052
  • I can follow every step, but in the end your answer is different from the one from Maxima. I have tried some values for x and the solutions are not equal. – marius Apr 09 '16 at 08:25
  • @ghostgate: Well, it's an indefinite integral, it's only defined up to an additive constant (which both you and I explicitly included in the result). So the question isn't whether the solutions are equal, but whether they differ only by an additive constant. – joriki Apr 09 '16 at 09:13
  • I have tried to solve definite integrals, like F(3)-F(2) and then the solution should be equal, or? – marius Apr 09 '16 at 09:16
  • @ghostgate: That integrates over the singularity at $x=\frac34\pi$. Note that we both didn't include the absolute value that should be around the argument of the logarithm; that doesn't matter as long as you don't try to integrate over a singularity where the sign of the argument changes. – joriki Apr 09 '16 at 09:24
  • 1
    Maybe I did a mistake while calculating, I will try a second time. Your way was the more elegant for me and the shortest. – marius Apr 09 '16 at 09:27
  • 1
    I have proved that your solution can be written as the one von Maxima, so I must have done a mistake. Thanks – marius Apr 09 '16 at 12:56
2

$\sin(x)+\cos(x)=\sqrt2\cos\left(x-\frac\pi4\right)$. Let $u=x-\frac\pi4$, then $$ \begin{align} \frac{\sin^2(x)\cos(x)}{\sin(x)+\cos(x)} &=\frac{(\cos(u)+\sin(u))^2}2\frac{\cos(u)-\sin(u)}{\sqrt2}\frac1{\sqrt2\cos(u)}\\ &=\frac{\cos(u)-\sin(u)+2\sin(u)\cos^2(u)-2\sin^2(u)\cos(u)}{4\cos(u)}\\ &=\frac14\left(1-\frac{\sin(u)}{\cos(u)}+2\sin(u)\cos(u)-2\sin^2(u)\right)\\ &=\frac14\left(\sin(2u)+\cos(2u)-\frac{\sin(u)}{\cos(u)}\right) \end{align} $$ Integrating gives $$ \begin{align} \int\frac{\sin^2(x)\cos(x)}{\sin(x)+\cos(x)}\mathrm{d}x+C &=-\frac18\cos(2u)+\frac18\sin(2u)+\frac14\log(\cos(u))+C\\ &=-\frac18\cos\left(2x-\frac\pi2\right)+\frac18\sin\left(2x-\frac\pi2\right)+\frac14\log\left(\cos\left(x-\frac\pi4\right)\right)+C\\ &=-\frac18\sin(2x)-\frac18\cos(2x)+\frac14\log(\cos(x)+\sin(x))+C-\frac{\log(2)}8 \end{align} $$

robjohn
  • 345,667