Evaluate the integral. ∫sin⁡hx1+cos⁡hxdx

Question

Evaluate the integral.  ∫sin⁡hx1+cos⁡hxdx

peterpan7117i · Accepted Answer

Step 1
Given data:
The given integral is:

\int \frac{\sin h x}{1 + \cos h x} d x

Assume

1 + \cos h x = u

and differentiate it with respect to x.

(0 + \sin h x) d x = d u

d x = \frac{d u}{\sin h x}

Step 2
Substitute the above-calculated values in the given expression.

\int \frac{\sin h x}{1 + \cos h x} d x = \int \frac{\sin h x}{u} (\frac{d u}{\sin h x})

= \int \frac{d u}{u}

= \ln (u) + C

Substitute

1 + \cos h x

for u in the above expression.

\int \frac{\sin h x}{1 + \cos h x} d x = \ln (1 + \cos h x) + C

Thus, the integral of the given expression is

\ln (1 + \cos h x) + C

.

intacte87 · Accepted Answer

Vasquez · Accepted Answer

Need to calculate: ∫hsin⁡(x)hcos⁡(x)+1dx=−∫1udu∫1udu=ln⁡(u)−∫1udu=−ln⁡(u)=−ln⁡(hcos⁡(x)+1)Answer:=−ln⁡(|hcos⁡(x)+1|)+C

Evaluate the integral. \int \frac{\sin h x}{1+\cos h x}dx