$\int_{a}^{b}\left(\frac{\cos(x)\tan^{\pi}(x)}{\sin^3(x)}\right)dx$

Question

I've got an integration problem; I don't know how to go from the left 'red' side to the right. Can someone help me?

Assuming $a<b$ and $a,b\in \mathbb{R}$:

$$\int_{a}^{b}\left(\frac{\cos(x)\tan^{\pi}(x)}{\sin^3(x)}\right)dx=\int_{a}^{b}\left(\frac{1}{\sin^3(x)}\cdot \left(\cos(x)\tan^{\pi}(x)\right)\right)dx=$$ $$\int_{a}^{b}\left(\csc^3(x)\cdot \left(\cos(x)\tan^{\pi}(x)\right)\right)dx=\int_{a}^{b}\left(\csc^3(x)\cos(x)\tan^{\pi}(x)\right)dx=$$ $$\int_{a}^{b}\left(\csc^2(x)\cot(x)\tan^{\pi}(x)\right)dx=\int_{a}^{b}\left(\tan^{\pi-1}\left(x\right)\csc^2(x)\right)dx$$

$$\color{red}{\int_{a}^{b}\left(\tan^{\pi-1}\left(x\right)\csc^2(x)\right)dx=\left[\frac{\tan^{\pi-2}(x)}{\pi-2}\right]_{a}^{b}}=$$

$$\left(\frac{\tan^{\pi-2}(b)}{\pi-2}\right)-\left(\frac{\tan^{\pi-2}(a)}{\pi-2}\right)=\frac{\tan^{\pi-2}(b)-\tan^{\pi-2}(a)}{\pi-2}$$

$\int { \tan ^{ \pi -1 }{ (x) } } \csc ^{ 2 }{ (x) } dx\quad \neq \quad \frac { \tan ^{ \pi -2 }{ (x) } }{ \pi -2 } $ — Oussama Boussif, Aug 05 '15 at 13:35

score 4 · Accepted Answer · answered Aug 05 '15 at 13:34

Replace $\tan$ with $\cot$:

\begin{align} \int_a^b \tan^u x \csc^2 x\,dx &= \int_a^b \cot^{-u} x \csc^2 x\,dx\\ &= -\int_a^b \cot^{-u} x \cot' x \,dx\\ &= -\biggl[\frac{\cot^{1-u} x}{1-u}\biggr]_a^b\\ &= \biggl[ \frac{\tan^{u-1} x}{u-1}\biggr]_a^b. \end{align}

score 2 · Answer 2 · answered Aug 05 '15 at 13:35

$$ \frac{\tan^{\beta-1}(x)}{\sin^2(x)} = \frac{\tan^2(x)}{\sin^2(x)}\tan^{\beta-3}(x) $$ this leads to $$ \int_a^b\frac{\tan^2(x)}{\sin^2(x)}\tan^{\beta-3}(x)dx = \int_a^b\sec^2 x\tan^{\beta-3}(x)dx $$ we have $$ \dfrac{d}{dx}\tan^\alpha x = \sec^2 x\tan^{\alpha-1}x $$

$\int_{a}^{b}\left(\frac{\cos(x)\tan^{\pi}(x)}{\sin^3(x)}\right)dx$

2 Answers2