Question

Evaluate: $\int \frac{tan\theta +ta{n}^3\theta }{1+ta{n}^3\theta }d\theta$

Answer 1

$\int \frac{\tan \theta +{\tan }^3\theta }{1+{\tan }^3\theta }d\theta =\int \frac{\tan \theta (1+{\tan }^3\theta )}{1+{\tan }^3\theta }d\theta$

Assuming $\tan \theta =x$

${\sec }^2\theta d\theta =dx$

So, $\int \frac{xdx}{1+x^3}\Rightarrow$

$\int \frac{x}{1+x^3}dx=\int \frac{-1dx}{3(x+1)}+\int \frac{1}{3}\frac{(x+1).dx}{x^2-x+1}$ $=\frac{1}{3}\left[\int \frac{2x-1}{x^2-x+1}dx+\int \frac{\frac{3}{2}dx}{x^2-x+1}\right]$

$=\frac{1}{3}[\frac{1}{2}\log (x^2-x+1)+\frac{3}{2}\int \frac{dx}{{(x-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}]$ [ we know: $\frac{dx}{x^2+a^2}={\tan }^{-1}\left(\frac{x}{a}\right)$

$=\frac{1}{3}[\frac{1}{2}\log (x^2-x+1)+\frac{3.2}{2\sqrt{3}}{\tan }^{-1}\frac{(x-\frac{1}{2})}{\left(\frac{\sqrt{3}}{2}\right)}]=\frac{1}{6}\log (x^2-x+1)+\sqrt{3}{\tan }^{-1}\frac{(x-\frac{1}{2})}{\left(\frac{\sqrt{3}}{2}\right)}$

&

$\tan -\frac{1}{3}\int \frac{dx}{x+1}=-\log (x+1)+C$

Partial fraction method

$\frac{x}{1+x^3}=\frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}$

$\frac{x}{1+x^3}=\frac{A(x^2-x+1)+(Bx+C)(x+1)}{(x+1)(x^2-x+1)}$

$A+B=0,A+C=0$

$C+B-A=1$

$A=\frac{-1}{3},B=C=\frac{1}{3}$