What is the integral of 1/(sin(t)) ?

The integral of 1/(sin(t)) is ln|tan(t/2)|+C

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integral of 1/(sin(t))

\int \frac{1}{sin ( t )} d t

ln tan (\frac{t}{2}) + C

Solution steps

\int \frac{1}{sin ( t )} d t

Apply u-substitution

: \int \frac{1}{u} d u

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = tan (\frac{t}{2})

= ln tan (\frac{t}{2})

Add a constant to the solution

= ln tan (\frac{t}{2}) + C

\int (2 x - 3 x^{2})^{- 0.5} d x

\int_{0}^{\infty} y^{2} e^{- y} d y

\int \frac{x ^{4}}{x ^{3} + x ^{2} - x - 1} d x

\int \frac{s}{2 s + 3} d s

f (x) = 6 + 5 x^{2} - 2 x^{3}, at x = 2

What is the integral of 1/(sin(t)) ?
The integral of 1/(sin(t)) is ln|tan(t/2)|+C