What is the integral of sin(41x)sec^2(cos(41x)) ?

The integral of sin(41x)sec^2(cos(41x)) is -1/41 tan(cos(41x))+C

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integral of sin(41x)sec^2(cos(41x))

\int sin (41 x) sec^{2} (cos (41 x)) d x

- \frac{1}{41} tan (cos (41 x)) + C

Solution steps

\int sin (41 x) sec^{2} (cos (41 x)) d x

Apply u-substitution

: \int sin (u) sec^{2} (cos (u)) \frac{1}{41} d u

= \int sin (u) sec^{2} (cos (u)) \frac{1}{41} d u

Take the constant out:

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

= \frac{1}{41} \cdot \int sin (u) sec^{2} (cos (u)) d u

Apply u-substitution

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

= \frac{1}{41} \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}{41} (- \int sec^{2} (v) d v)

Use the common integral:

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

= \frac{1}{41} (- tan (v))

Substitute back

= \frac{1}{41} (- tan (cos (41 x)))

Simplify

= - \frac{1}{41} tan (cos (41 x))

Add a constant to the solution

= - \frac{1}{41} tan (cos (41 x)) + C

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What is the integral of sin(41x)sec^2(cos(41x)) ?
The integral of sin(41x)sec^2(cos(41x)) is -1/41 tan(cos(41x))+C