Answer:
ln|sec θ + tan θ| + C
Step-by-step explanation:
The integrals of basic trig functions are:
∫ sin θ dθ = -cos θ + C
∫ cos θ dθ = sin θ + C
∫ csc θ dθ = -ln|csc θ + cot θ| + C
∫ sec θ dθ = ln|sec θ + tan θ| + C
∫ tan θ dθ = -ln|cos θ| + C
∫ cot θ dθ = ln|sin θ| + C
The integral of sec θ can be proven by multiplying and dividing by sec θ + tan θ, then using ∫ du/u = ln|u| + C.
∫ sec θ dθ
∫ sec θ (sec θ + tan θ) / (sec θ + tan θ) dθ
∫ (sec² θ + sec θ tan θ) / (sec θ + tan θ) dθ
ln|sec θ + tan θ| + C
<span>Set up the long division.
</span>
<span>6|15</span>
<span>Calculate 15 ÷ 6, which is 2 with a remainder of 3.
</span>
<span>2<span>6|15</span><span>12</span>3</span>
<span>
Answer
</span>2<span> with a remainder of </span>3
<span>Answer (Mixed Fraction)
</span><span>2</span>

5 <span>Simplify </span>

<span> to </span>

<span>
</span><span>2</span>

Done
<span>Decimal Form: 2.5</span>
4+6=10 6/10 60% that is the answer