I=∫ sec^3 θ d θ =sec θ∫ sec^2 θ dθ ∫ tan θ (sec θ tan θ) d θ =sec θ tan θ ∫ (sec^2 θ 1)sec θ dθ =sec θ tan θ 1 + ln |sec θ+tan θ| ⇒ I = 1/2[sec θ tan ...

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Byju's Answer

Standard XII

Mathematics

Integration of Trigonometric Functions

Evaluate ∫ 3θ...

Question

Evaluate $\int s e c^{3} θ d θ$

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Solution

$I = \int s e c^{3} θ d θ = s e c θ \int s e c^{2} θ d θ - \int t a n θ (s e c θ t a n θ) d θ = s e c θ t a n θ - \int (s e c^{2} θ - 1) s e c θ d θ = s e c θ t a n θ - 1 + l n | s e c θ + t a n θ | \Rightarrow I = \frac{1}{2} [s e c θ t a n θ] + \frac{1}{2} l n | s e c θ t a n θ | + c$

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Evaluate ∫sec3θ dθ

I=∫sec3θ dθ=sec θ∫sec2 θ dθ−∫tan θ(sec θ tan θ) dθ=sec θ tan θ−∫(sec2 θ−1)sec θ dθ=sec θ tan θ−1+ln|sec θ+tan θ|⇒I=12[sec θ tan θ]+12 ln |sec θ tan θ|+c

Evaluate $\int s e c^{3} θ d θ$