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User Gypaetus
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1 Answer

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Answer:

(e) csc x − cot x − ln(1 + cos x) + C

(c) 0

Explanation:

(e) ∫ (1 + sin x) / (1 + cos x) dx

Split the integral.

∫ 1 / (1 + cos x) dx + ∫ sin x / (1 + cos x) dx

Multiply top and bottom of first integral by the conjugate, 1 − cos x.

∫ (1 − cos x) / (1 − cos²x) dx + ∫ sin x / (1 + cos x) dx

Pythagorean identity.

∫ (1 − cos x) / (sin²x) dx + ∫ sin x / (1 + cos x) dx

Divide.

∫ (csc²x − cot x csc x) dx + ∫ sin x / (1 + cos x) dx

Integrate.

csc x − cot x − ln(1 + cos x) + C

(c) ∫₋₇⁷ erf(x) dx

= ∫₋₇⁰ erf(x) dx + ∫₀⁷ erf(x) dx

The error function is odd (erf(-x) = -erf(x)), so:

= -∫₀⁷ erf(x) dx + ∫₀⁷ erf(x) dx

= 0

User EGeuens
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