What is the integral of na^{cos(bt)}sin(bt) ?

The integral of na^{cos(bt)}sin(bt) is -na^{cos(bt)} 1/(bln(a))+C

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integral of na^{cos(bt)}sin(bt)

\int n a^{c o s (b t)} sin (b t) d t

- n a^{c o s (b t)} \frac{1}{b ln ( a )} + C

Solution steps

\int n a^{c o s (b t)} sin (b t) d t

Take the constant out:

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

= n \cdot \int sin (b t) a^{c o s (b t)} d t

Apply u-substitution

: \int - \frac{1}{b ln ( a )} d u

= n \cdot \int - \frac{1}{b ln ( a )} d u

Integral of a constant:

\int a d x = a x

= n (- \frac{1}{b ln ( a )}) u

Substitute back

u = a^{c o s (b t)}

= n (- \frac{1}{b ln ( a )}) a^{c o s (b t)}

Simplify

= - n a^{c o s (b t)} \frac{1}{b ln ( a )}

Add a constant to the solution

= - n a^{c o s (b t)} \frac{1}{b ln ( a )} + C

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What is the integral of na^{cos(bt)}sin(bt) ?
The integral of na^{cos(bt)}sin(bt) is -na^{cos(bt)} 1/(bln(a))+C