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

The integral of sin(41x)sin(cos(41x)) is 1/41 cos(cos(41x))+C

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integral of sin(41x)sin(cos(41x))

\int sin (41 x) sin (cos (41 x)) d x

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

Solution steps

\int sin (41 x) sin (cos (41 x)) d x

Apply u-substitution

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

= \int sin (u) sin (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) sin (cos (u)) d u

Apply u-substitution

: \int - sin (v) d v

= \frac{1}{41} \cdot \int - sin (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 sin (v) d v)

Use the common integral:

\int sin (v) d v = - cos (v)

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

Substitute back

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

Simplify

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

Add a constant to the solution

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

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What is the integral of sin(41x)sin(cos(41x)) ?
The integral of sin(41x)sin(cos(41x)) is 1/41 cos(cos(41x))+C