Question

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

• A. $-\frac{1}{3}log\left|1+tan\theta \right|-\frac{1}{6}log\left|ta{n}^2\theta -tan\theta +1\right|-\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{2tan\theta -1}{\sqrt{3}}\right)+C$
• B. $-\frac{1}{3}log\left|1+tan\theta \right|+\frac{1}{6}log\left|ta{n}^2\theta -tan\theta +1\right|+\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{2tan\theta -1}{\sqrt{3}}\right)+C$
• C. $-\frac{1}{3}log\left|1+tan\theta \right|+\frac{1}{6}log\left|ta{n}^2\theta +tan\theta +1\right|-\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{2tan\theta +1}{\sqrt{3}}\right)+C$
• D. None of these

Answer 1

Answer: (B)

We have,

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

$\int \frac{tan\theta se{c}^2\theta }{\left(1+ta{n}^3\theta \right)}d\theta$

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

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

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 $B=-A=\frac{1}{3}andC=-A=\frac{1}{3}$

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

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

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

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

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

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

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