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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int n a^{s i n (b x)} cos (b x) d x

\frac{n a ^{s i n (b x)}}{b ln ( a )} + C

Solution steps

\int n a^{s i n (b x)} cos (b x) d x

Take the constant out:

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

= n \cdot \int cos (b x) a^{s i n (b x)} d x

Apply u-substitution

: \int \frac{a ^{s i n (u)} cos ( u )}{b} d u

= n \cdot \int \frac{a ^{s i n (u)} cos ( u )}{b} d u

Take the constant out:

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

= n \frac{1}{b} \cdot \int a^{s i n (u)} cos (u) d u

Apply u-substitution

: \int a^{v} d v

= n \frac{1}{b} \cdot \int a^{v} d v

\int a^{x} d x = \frac{a ^{x}}{ln a}

= n \frac{1}{b} \cdot \frac{a ^{v}}{ln ( a )}

Substitute back

= n \frac{1}{b} \cdot \frac{a ^{s i n (b x)}}{ln ( a )}

Simplify

n \frac{1}{b} \cdot \frac{a ^{s i n (b x)}}{ln ( a )} : \frac{n a ^{s i n (b x)}}{b ln ( a )}

= \frac{n a ^{s i n (b x)}}{b ln ( a )}

Add a constant to the solution

= \frac{n a ^{s i n (b x)}}{b ln ( a )} + C

\int \frac{x + 2}{x ^{2} + 2 x + 3} d x

\int \frac{e ^{x}}{( e ^{x} + 1 ) ^{13}} d x

\int \frac{3 x ^{b - 1}}{a + x ^{b}} d x

\int ln (x e^{x}) d x

\int \frac{x ^{2}}{1 - 4 x ^{2}} d x

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