Question Number 19077 by Joel577 last updated on 04/Aug/17
$$\int\:\mathrm{sec}^{\mathrm{3}} \:{x}\:{dx} \\ $$
Answered by ajfour last updated on 04/Aug/17
![I=∫sec θ(sec^2 θdθ) =sec θ∫sec^2 θdθ−∫[(d/dθ)(sec θ).∫sec^2 θdθ]dθ =sec θtan θ−∫(sec θtan θ)tan θdθ+C_1 =sec θtan θ−∫(sec θ)(sec^2 θ−1)dθ+C_1 I=sec θtan θ−∫sec^3 θdθ+∫sec θdθ+C_1 I=sec θtan θ−I+ln ∣sec θ+tan θ∣+C_1 +C_2 ⇒ I=(1/2)sec θtan θ +(1/2)ln ∣sec θ+tan θ∣+C .](https://www.tinkutara.com/question/Q19086.png)
$$\mathrm{I}=\int\mathrm{sec}\:\theta\left(\mathrm{sec}\:^{\mathrm{2}} \theta\mathrm{d}\theta\right) \\ $$$$=\mathrm{sec}\:\theta\int\mathrm{sec}\:^{\mathrm{2}} \theta\mathrm{d}\theta−\int\left[\frac{\mathrm{d}}{\mathrm{d}\theta}\left(\mathrm{sec}\:\theta\right).\int\mathrm{sec}\:^{\mathrm{2}} \theta\mathrm{d}\theta\right]\mathrm{d}\theta \\ $$$$=\mathrm{sec}\:\theta\mathrm{tan}\:\theta−\int\left(\mathrm{sec}\:\theta\mathrm{tan}\:\theta\right)\mathrm{tan}\:\theta\mathrm{d}\theta+\mathrm{C}_{\mathrm{1}} \\ $$$$=\mathrm{sec}\:\theta\mathrm{tan}\:\theta−\int\left(\mathrm{sec}\:\theta\right)\left(\mathrm{sec}\:^{\mathrm{2}} \theta−\mathrm{1}\right)\mathrm{d}\theta+\mathrm{C}_{\mathrm{1}} \\ $$$$\mathrm{I}=\mathrm{sec}\:\theta\mathrm{tan}\:\theta−\int\mathrm{sec}\:^{\mathrm{3}} \theta\mathrm{d}\theta+\int\mathrm{sec}\:\theta\mathrm{d}\theta+\mathrm{C}_{\mathrm{1}} \\ $$$$\mathrm{I}=\mathrm{sec}\:\theta\mathrm{tan}\:\theta−\mathrm{I}+\mathrm{ln}\:\mid\mathrm{sec}\:\theta+\mathrm{tan}\:\theta\mid+\mathrm{C}_{\mathrm{1}} +\mathrm{C}_{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{I}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}\:\theta\mathrm{tan}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{sec}\:\theta+\mathrm{tan}\:\theta\mid+\mathrm{C}\:. \\ $$
Commented by Joel577 last updated on 04/Aug/17
$${thank}\:{you}\:{very}\:{much} \\ $$