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  4. Evaluate:∫(tan θ +tan3θ/1+tan3 θ) dθ

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Q. Evaluate:$\int\frac{tan\, \theta +tan^{3}\theta}{1+tan^{3} \theta} d\theta $

Integrals

Solution:

We have,

$\int\frac{tan\, \theta +tan^{3}\theta}{1+tan^3 \theta} d\theta $

$\int\frac{tan\, \theta sec^{2}\theta}{\left(1+tan^{3 }\theta\right)}d\theta $

Putting$ tan\, \theta =t \Rightarrow sec^{2}\theta d\theta = dt$, we get

$ I= \int \frac{t dt}{\left(1+t^{3}\right)} = \int\frac{t}{\left(1+t\right)\left(t^{2}-t +1\right)}dt$

Let $\frac{t}{\left(1+t\right)\left(1-t+t^{2}\right) }= \frac{A}{1+t} +\frac{Bt +C}{1-t+t^{2}}$

$\Rightarrow t= A\left(1-t+t^{2}\right) +\left(Bt+C\right)\left(t+1\right) ............\left(i\right)$

put $t=-1in \left(i\right)$, we get $A= -\frac{1}{3}$

Comparing the coefficients of $t^{2}$ and constant terms on both sides of $\left(i\right)$, we get $ B= -A = \frac{1}{3}and C= -A =\frac{1}{3}$

$\Rightarrow I= -\frac{1}{3}\int\frac{1}{1+t}dt +\frac{1}{3}\frac{ t+1}{t^{2}-t+1}dt$

$\Rightarrow I= -\frac{1}{3} \int\frac{t+1}{t^{2}-t+1} dt$

$\Rightarrow I= -\frac{1}{3}\int\frac{1}{1+t}dt +\frac{1}{6}\int\frac{2t-1+3}{t^{2}-t+1} dt $

$\Rightarrow I= -\frac{1}{3}\int\frac{1}{1+t}dt +\frac{1}{6}\int\frac{2t-1}{t^{2}-t+1} dt +\frac{1}{2}\int\frac{1}{\left(t-\frac{1}{2}\right)^{2}+\left(\sqrt{\frac{3}{2}}\right)^{2}}dt$

$\Rightarrow I= -\frac{1}{3}log\left|1+t\right| +\frac{1}{6}log\left|t^{2}-t +1\right|+\frac{1}{2}\cdot\frac{1}{\frac{\sqrt{3}}{2}} tan ^{-1}\left(\frac{t-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C$

$ \Rightarrow I= -\frac{1}{3}log\left|1+t\right| +\frac{1}{6}log\left|t^{2}-t+1\right|+\frac{1}{\sqrt{3}}tan^{-1}\left(\frac{2t -1}{\sqrt{3}}\right)$

$\therefore I= -\frac{1}{3}log\left|1+tan \theta\right| +\frac{1}{6 }log\left|tan^{2}\theta-tan\theta +1\right| +\frac{1}{\sqrt{3}}tan^{-1}\left(\frac{2tan\theta -1}{\sqrt{3}}\right) +C$