sec-3-x-dx-

Question Number 19077 by Joel577 last updated on 04/Aug/17

\int \sec^{3} x d x

Answered by ajfour last updated on 04/Aug/17

I = \int \sec θ (\sec^{2} θ d θ)

= \sec θ \int \sec^{2} θ d θ - \int [\frac{d}{d θ} (\sec θ) . \int \sec^{2} θ d θ] d θ

= \sec θ \tan θ - \int (\sec θ \tan θ) \tan θ d θ + C_{1}

= \sec θ \tan θ - \int (\sec θ) (\sec^{2} θ - 1) d θ + C_{1}

I = \sec θ \tan θ - \int \sec^{3} θ d θ + \int \sec θ d θ + C_{1}

I = \sec θ \tan θ - I + \ln ∣ \sec θ + \tan θ ∣ + C_{1} + C_{2}

\Rightarrow I = \frac{1}{2} \sec θ \tan θ

+ \frac{1}{2} \ln ∣ \sec θ + \tan θ ∣ + C .

Commented by Joel577 last updated on 04/Aug/17

t h a n k y o u v e r y m u c h

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