What is the integral of sin(18t)sec^2(cos(18t)) ?

The integral of sin(18t)sec^2(cos(18t)) is -1/18 tan(cos(18t))+C

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integral of sin(18t)sec^2(cos(18t))

\int sin (18 t) sec^{2} (cos (18 t)) d t

- \frac{1}{18} tan (cos (18 t)) + C

Solution steps

\int sin (18 t) sec^{2} (cos (18 t)) d t

Apply u-substitution

: \int sin (u) sec^{2} (cos (u)) \frac{1}{18} d u

= \int sin (u) sec^{2} (cos (u)) \frac{1}{18} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{18} \cdot \int sin (u) sec^{2} (cos (u)) d u

Apply u-substitution

: \int - sec^{2} (v) d v

= \frac{1}{18} \cdot \int - sec^{2} (v) d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{18} (- \int sec^{2} (v) d v)

Use the common integral:

\int sec^{2} (v) d v = tan (v)

= \frac{1}{18} (- tan (v))

Substitute back

= \frac{1}{18} (- tan (cos (18 t)))

Simplify

= - \frac{1}{18} tan (cos (18 t))

Add a constant to the solution

= - \frac{1}{18} tan (cos (18 t)) + C

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What is the integral of sin(18t)sec^2(cos(18t)) ?
The integral of sin(18t)sec^2(cos(18t)) is -1/18 tan(cos(18t))+C