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Question: \[\int_{}^{}\frac{\mathbf{d\theta}}{\mathbf{\sin}\mathbf{\theta}\mathbf{.}\mathbf{\cos}^{\mathbf{3}}...

∫dθsin⁡θ.cos⁡3θ=\int_{}^{}\frac{\mathbf{d\theta}}{\mathbf{\sin}\mathbf{\theta}\mathbf{.}\mathbf{\cos}^{\mathbf{3}}\mathbf{\theta}}\mathbf{=}

A

logtanθ+tan⁡2θ+c{logtan}\theta + \tan^{2}\theta + c

B

logtanθ−12tan⁡2θ+c{logtan}\theta - \frac{1}{2}\tan^{2}\theta + c

C

logtanθ+12tan⁡2θ+c{logtan}\theta + \frac{1}{2}\tan^{2}\theta + c

D

None of these

Answer

logtanθ+12tan⁡2θ+c{logtan}\theta + \frac{1}{2}\tan^{2}\theta + c

Explanation

Solution

∫dθsin⁡θcos⁡3θ=∫sec⁡2θdθsin⁡θcos⁡θ=∫sec⁡2θ(1+tan⁡2θ)dθtan⁡θdθ\int_{}^{}\frac{d\theta}{\sin\theta\cos^{3}\theta} = \int_{}^{}{\frac{\sec^{2}\theta d\theta}{\sin\theta\cos\theta} = \int_{}^{}\frac{\sec^{2}\theta(1 + \tan^{2}\theta)d\theta}{\tan\theta d\theta}}Put t=tan⁡θt = \tan\theta ⇒dt=sec⁡2θdθ\Rightarrow dt = \sec^{2}\theta d\theta then it reduces to

⇒\Rightarrow ∫1+t2t⥂⥂dt=∫(1t+t)dt\int_{}^{}{\frac{1 + t^{2}}{t} ⥂ ⥂ dt} = \int_{}^{}\left( \frac{1}{t} + t \right)dt =log⁡t+t22+c=log⁡tan⁡θ+12tan⁡2θ+c.= \log t + \frac{t^{2}}{2} + c = \log{\tan\theta} + \frac{1}{2}\tan^{2}\theta + c.