Show that $\int_{0}^{\pi}{\sin^5(x)\over 1+\cos^3(x)}\mathrm dx =\ln{3}?$

Question

$$\int_{0}^{\pi}{\sin^5(x)\over 1+\cos^3(x)}\mathrm dx =\ln{3}\tag1$$

How can we show that $(1)=\ln{3}$

$u=\sin^3{x}$ then $du=3\sin^2{x}\cos{x}dx$

$${1\over 3}\int{u\mathrm du\over \cos{x}+\cos^4{x}}\tag2$$

This is not a good substitution

The solution to this problem already exists here – Your IDE Dec 09 '17 at 20:11 — Your IDE, Dec 09 '17 at 20:11
If you are ok, you can set as solved. Thanks! – user Dec 13 '17 at 07:23 — user, Dec 13 '17 at 07:23

score 3 · Accepted Answer · answered Dec 09 '17 at 20:02

$$\int_0^\pi \frac{\sin^5 x}{1+\cos^3 x}dx$$ $$=\int_0^\pi \frac{-(1-\cos^2 x)^2}{1+\cos^3 x}\cdot -\sin x\space dx$$ Now let $u=\cos x$. Then we have $$=\int_1^{-1} \frac{-(1-u^2)^2}{1+u^3}\cdot du$$ $$=\int_{-1}^{1} \frac{(1-u^2)^2}{1+u^3}\cdot du$$ Can you finish this?

score 2 · Answer 2 · answered Dec 09 '17 at 20:01

2

HINT

Note that: $$sin^5(x)=\sin^2 x \sin^2 x \sin x=(1-cos^2 x)^2 \sin x$$

This suggest to set: $\cos x =u$

answered Dec 09 '17 at 20:01

user

154,566

score 1 · Answer 3 · answered Dec 09 '17 at 20:04

1

Try: $$ u = \cos(x) \\ du = -\sin(x)dx $$ Then $$\frac{sin^5(x)}{1+\cos^3(x)}dx = -\frac{(1-u^2)^2}{1+u^3}du$$

answered Dec 09 '17 at 20:04

mate89

538

score 1 · Answer 4 · answered Dec 09 '17 at 20:19

Without using integration by substitution or by parts:

Trigonometric algebra

We use the identity that $(1-\cos x)(1+ \cos^3x)=(1-\cos^2x)(1-\cos x+\cos^2x)$ so $$\frac{\sin^2x}{1+\cos^3x}=\frac{1-\cos x}{1-\cos x + \cos^2x} \implies \frac{\sin^5x}{1+\cos^3x}=\frac{\sin^3x(1-\cos x)}{1-\cos x + \cos^2x}$$ Now RHS is equivalent to $$\frac{\sin x(1-\cos^2x)(1-\cos x)}{1-\cos x + \cos^2x}=\frac{\sin x-\sin x \cos x+\sin x \cos^2x(\cos x-1)}{1-\cos x + \cos^2x}$$ or $$\frac{\sin x(1-2\cos x)+\sin x \cos x(1-\cos x + \cos^2x)}{1-\cos x + \cos^2x}=\frac{2\sin x(1-\cos^2x)}{2-2\cos x + 2\cos^2x}+\sin x \cos x$$ We do this so that we can replace the square term with $\cos 2x$. Continuing, we have $$\frac{2\sin x - 4 \sin x \cos x}{3-2\cos x + \cos 2x}+\frac12 \sin 2x=\frac{-2(-\sin x)+2(-\sin 2x)}{3-2\cos x + \cos 2x}-\frac12(-\sin 2x)$$ Notice that the numerator is the derivative of the denominator! Thus the integral is $$\int \frac{\sin^5x}{1+\cos^3x} dx=\ln|3-2\cos x + \cos 2x|-\frac14\cos 2x + c$$ where $c$ is a constant.

Substituting the upper and lower limits of $\pi$ and $-\pi$ respectively yields $\ln 3$.

score 1 · Answer 5 · answered Dec 09 '17 at 20:23

Do you know about the Bioche's rules for rational functions of $\cos,\sin$.

Wiki : Règles de Bioche

In our case $w(x)=\dfrac{\sin(x)^5}{1+\cos(x)^3}\mathop{dx}$ and we can see that $w(-x)=w(x)$

Thus a substitution $\boxed{u=\cos(x)}$ is appropriate.

After some calculation you get $\displaystyle I=\int_{-1}^1 \left(u-\dfrac{2u-1}{u^2-u+1}\right)\mathop{du}$ with a nice $\frac {f'}{f}$ appearance.

score 1 · Answer 6 · answered Dec 09 '17 at 20:25

$$\int_{0}^{\pi}\frac{\sin^5(x)}{1+\cos^3(x)}\,dx = \int_{0}^{\pi/2}\frac{\sin^5(x)}{1+\cos^3(x)}+\frac{\sin^5(x)}{1-\cos^3(x)}\,dx $$ equals: $$ \int_{0}^{\pi/2}\frac{2\sin^5(x)}{1-\cos(x)^6}\,dx = \int_{0}^{\pi/2}\frac{2\cos(x)^5}{1-\sin(x)^6}\,dx $$ or, by setting $\sin(x)=u$, $$ \int_{0}^{1}\frac{2\sqrt{1-u^2}^4}{1-u^6}\,du=2\int_{0}^{1}\frac{1-u^2}{1+u^2+u^4}\,du $$ which is an elementary integral (by partial fraction decomposition).

Show that $\int_{0}^{\pi}{\sin^5(x)\over 1+\cos^3(x)}\mathrm dx =\ln{3}?$

6 Answers6