What is the integral of sin(13t)sec^2(cos(13t)) ?

The integral of sin(13t)sec^2(cos(13t)) is -1/13 tan(cos(13t))+C

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integral of sin(13t)sec^2(cos(13t))

\int sin (13 t) sec^{2} (cos (13 t)) d t

- \frac{1}{13} tan (cos (13 t)) + C

Solution steps

\int sin (13 t) sec^{2} (cos (13 t)) d t

Apply u-substitution

: \int sin (u) sec^{2} (cos (u)) \frac{1}{13} d u

= \int sin (u) sec^{2} (cos (u)) \frac{1}{13} d u

Take the constant out:

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

= \frac{1}{13} \cdot \int sin (u) sec^{2} (cos (u)) d u

Apply u-substitution

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

= \frac{1}{13} \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}{13} (- \int sec^{2} (v) d v)

Use the common integral:

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

= \frac{1}{13} (- tan (v))

Substitute back

= \frac{1}{13} (- tan (cos (13 t)))

Simplify

= - \frac{1}{13} tan (cos (13 t))

Add a constant to the solution

= - \frac{1}{13} tan (cos (13 t)) + C

\int (x^{2} + 1) (2 x) d x

\int 2 x + \frac{5}{2 x ^{2}} d x

\int \frac{1}{v ( v + b )} d v

\int \frac{e ^{x}}{- 3 + e ^{x}} d x

cos^{2} (2 x)

What is the integral of sin(13t)sec^2(cos(13t)) ?
The integral of sin(13t)sec^2(cos(13t)) is -1/13 tan(cos(13t))+C