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Question: \[\int_{}^{}\frac{\mathbf{\tan}\mathbf{\varphi}\mathbf{+}\mathbf{\tan}^{\mathbf{3}}\mathbf{\varphi}}...

∫tan⁡φ+tan⁡3φ1+tan⁡3φdφ\int_{}^{}\frac{\mathbf{\tan}\mathbf{\varphi}\mathbf{+}\mathbf{\tan}^{\mathbf{3}}\mathbf{\varphi}}{\mathbf{1 +}\mathbf{\tan}^{\mathbf{3}}\mathbf{\varphi}}\mathbf{d\varphi}

A

13log⁡∣1+tan⁡φ∣+16log⁡∣tan⁡2φ−tan⁡φ+1∣+13tan⁡−1(2tan⁡φ−13)+c\frac{1}{3}\log|1 + \tan\varphi| + \frac{1}{6}\log|\tan^{2}\varphi - \tan\varphi + 1| + \frac{1}{3}\tan^{- 1}\left( \frac{2\tan\varphi - 1}{\sqrt{3}} \right) + c

B

−13log⁡∣1+tan⁡φ∣+16log⁡∣tan⁡2φ−tan⁡φ+1∣+13tan⁡−1(2tan⁡φ−13)+c\frac{- 1}{3}\log|1 + \tan\varphi| + \frac{1}{6}\log|\tan^{2}\varphi - \tan\varphi + 1| + \frac{1}{\sqrt{3}}\tan^{- 1}\left( \frac{2\tan\varphi - 1}{\sqrt{3}} \right) + c

C

13log⁡(1+tan⁡2φtan⁡3φ)+c\frac{1}{3}\log\left( \frac{1 + \tan^{2}\varphi}{\tan^{3}\varphi} \right) + c

D

None of these

Answer

−13log⁡∣1+tan⁡φ∣+16log⁡∣tan⁡2φ−tan⁡φ+1∣+13tan⁡−1(2tan⁡φ−13)+c\frac{- 1}{3}\log|1 + \tan\varphi| + \frac{1}{6}\log|\tan^{2}\varphi - \tan\varphi + 1| + \frac{1}{\sqrt{3}}\tan^{- 1}\left( \frac{2\tan\varphi - 1}{\sqrt{3}} \right) + c

Explanation

Solution

Let I=∫tan⁡φ+tan⁡3φ1+tan⁡3φdφ=∫tan⁡φ(1+tan⁡2φ)1+tan⁡3φI = \int_{}^{}\frac{\tan\varphi + \tan^{3}\varphi}{1 + \tan^{3}\varphi}d\varphi = \int_{}^{}\frac{\tan\varphi(1 + \tan^{2}\varphi)}{1 + \tan^{3}\varphi}

⇒\Rightarrow I=∫tan⁡φsec⁡2φ1+tan⁡3φdφI = \int_{}^{}\frac{\tan\varphi\sec^{2}\varphi}{1 + \tan^{3}\varphi}d\varphi

Putting tan⁡φ=t⇒sec⁡2⥂φdφ=dt,\tan\varphi = t \Rightarrow \sec^{2} ⥂ \varphi d\varphi = dt, we have,

I=∫t1+t3dt=∫tdt(1+t)(1−t+t2)I = \int_{}^{}{\frac{t}{1 + t^{3}}dt =}\int_{}^{}\frac{tdt}{(1 + t)(1 - t + t^{2})}Let t(1+t)(1−t+t2)=A1+t+Bt+C1−t+t2⇒\frac{t}{(1 + t)(1 - t + t^{2})} = \frac{A}{1 + t} + \frac{Bt + C}{1 - t + t^{2}} \Rightarrow t=A(1−t+t2)+(Bt+C)(1+t)t = A(1 - t + t^{2}) + (Bt + C)(1 + t) ......(i)

Putting t=−1t = - 1 in equation (i), we get,

−1=A(1+1+1)⇒A=−13- 1 = A(1 + 1 + 1) \Rightarrow A = - \frac{1}{3}

Comparing the coefficients of t2t^{2} and constant terms on both the sides of equation (i) we get

0=A+B⇒B=−A=130 = A + B \Rightarrow B = - A = \frac{1}{3} and 0=A+C⇒C=−A=130 = A + C \Rightarrow C = - A = \frac{1}{3}

∴\therefore t(1+t)(1−t+t2)=−13.11+r+13.t+11−t+t2\frac{t}{(1 + t)(1 - t + t^{2})} = - \frac{1}{3}.\frac{1}{1 + r} + \frac{1}{3}.\frac{t + 1}{1 - t + t^{2}}

Hence, I=−13∫dt1+t+13∫t+1t2−t+1dtI = - \frac{1}{3}\int_{}^{}{\frac{dt}{1 + t} + \frac{1}{3}\int_{}^{}{\frac{t + 1}{t^{2} - t + 1}dt}}

=−13∫dt1+t+16∫2t+2t2−t+1dt= - \frac{1}{3}\int_{}^{}{\frac{dt}{1 + t} + \frac{1}{6}\int_{}^{}{\frac{2t + 2}{t^{2} - t + 1}dt}} =−13∫dt1+t+16∫(2t−1)+3t2−t+1dt= - \frac{1}{3}\int_{}^{}{\frac{dt}{1 + t} + \frac{1}{6}\int_{}^{}{\frac{(2t - 1) + 3}{t^{2} - t + 1}dt}}

=−13∫dt1+t+16∫2t−1t2−t+1dt+36.∫dtt2−t+1= - \frac{1}{3}\int_{}^{}{\frac{dt}{1 + t} + \frac{1}{6}\int_{}^{}{\frac{2t - 1}{t^{2} - t + 1}dt + \frac{3}{6}.\int_{}^{}\frac{dt}{t^{2} - t + 1}}} =−13∫dt1+t+16∫2t−1t2−t+1dt+12∫dt(t−12)2+34=−13log⁡∣1+t∣+16log⁡∣t2−t+1∣+12.23tan⁡−1(t−1232)+c=−13log⁡∣1+t∣+16log⁡∣t2−t+1∣+13tan⁡−1(2t−13)+c= - \frac{1}{3}\int_{}^{}{\frac{dt}{1 + t} + \frac{1}{6}\int_{}^{}{\frac{2t - 1}{t^{2} - t + 1}dt + \frac{1}{2}\int_{}^{}\frac{dt}{\left( t - \frac{1}{2} \right)^{2} + \frac{3}{4}}}} = - \frac{1}{3}\log|1 + t| + \frac{1}{6}\log|t^{2} - t + 1| + \frac{1}{2}.\frac{2}{\sqrt{3}}\tan^{- 1}\left( \frac{t - \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) + c = - \frac{1}{3}\log|1 + t| + \frac{1}{6}\log|t^{2} - t + 1| + \frac{1}{3}\tan^{- 1}\left( \frac{2t - 1}{\sqrt{3}} \right) + c =−13log⁡∣1+tan⁡φ∣+16log⁡∣tan⁡2φ−tan⁡φ+1∣+13tan⁡−1(2tan⁡φ−13)+c.= - \frac{1}{3}\log|1 + \tan\varphi| + \frac{1}{6}\log|\tan^{2}\varphi - \tan\varphi + 1| + \frac{1}{\sqrt{3}}\tan^{- 1}\left( \frac{2\tan\varphi - 1}{\sqrt{3}} \right) + c.