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Question

Question: \[\int_{}^{}{\tan^{5}\theta d\theta =}\]...

∫tan⁡5θdθ=\int_{}^{}{\tan^{5}\theta d\theta =}

A

tan⁡4θ4−tan⁡2θ2+logsecθ+c\frac{\tan^{4}\theta}{4} - \frac{\tan^{2}\theta}{2} + {logsec}\theta + c

B

tan⁡4θ4−tan⁡2θ2−logsecθ+c\frac{\tan^{4}\theta}{4} - \frac{\tan^{2}\theta}{2} - {logsec}\theta + c

C

tan⁡4θ4−tan⁡2θ2+log⁡∣sec⁡θ∣+c\frac{\tan^{4}\theta}{4} - \frac{\tan^{2}\theta}{2} + \log|\sec\theta| + c

D

None of these

Answer

tan⁡4θ4−tan⁡2θ2+log⁡∣sec⁡θ∣+c\frac{\tan^{4}\theta}{4} - \frac{\tan^{2}\theta}{2} + \log|\sec\theta| + c

Explanation

Solution

∫tan⁡5θdθ=I5=tan⁡4θ4−I3\int_{}^{}{\tan^{5}\theta d\theta} = I_{5} = \frac{\tan^{4}\theta}{4} - I_{3}

=tan⁡4θ4−tan⁡2θ2+I1= \frac{\tan^{4}\theta}{4} - \frac{\tan^{2}\theta}{2} + I_{1} =tan⁡4θ4−tan⁡2θ2+log⁡∣sec⁡θ∣+c.= \frac{\tan^{4}\theta}{4} - \frac{\tan^{2}\theta}{2} + \log|\sec\theta| + c.