tan-3-sin-d-

Question Number 129946 by bemath last updated on 21/Jan/21

\int \frac{\tan φ + 3}{\sin φ} d φ ?

Answered by liberty last updated on 21/Jan/21

T = \int (\frac{1}{\cos φ} + \frac{3}{\sin φ}) d φ

T = \int (\sec φ + 3 \csc φ) d φ

T = \ln ∣ \sec φ + \tan φ ∣ - 3 \ln ∣ \cot φ + \csc φ ∣ + c

T = \ln ∣ \frac{1 + \sin φ}{\cos φ} ∣ - 3 \ln ∣ \frac{\cos φ + 1}{\sin φ} ∣ + c

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