What is the integral of e^{sin(x)+1}cos(x) ?

The integral of e^{sin(x)+1}cos(x) is e^{1+sin(x)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{sin(x)+1}cos(x)

\int e^{s i n (x) + 1} cos (x) d x

e^{1 + s i n (x)} + C

Solution steps

\int e^{s i n (x) + 1} cos (x) d x

e^{s i n (x) + 1} cos (x) = e^{1} e^{s i n (x)} cos (x)

= \int e^{1} e^{s i n (x)} cos (x) d x

Take the constant out:

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

= e^{1} \cdot \int e^{s i n (x)} cos (x) d x

Simplify

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

Apply u-substitution

: \int e^{u} d u

= e \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= e e^{u}

Substitute back

u = sin (x)

= e e^{s i n (x)}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

e e^{s i n (x)} = e^{1 + s i n (x)}

= e^{1 + s i n (x)}

Add a constant to the solution

= e^{1 + s i n (x)} + C

\int \frac{x ^{2} - 2 x + 1}{x ^{3} + x} d x

\int \frac{x + 3}{( x - 1 ) ( x + 2 ) ^{2}} d x

\int \frac{1}{1 - 0.4 v} d v

\int (2 x sin (x)) d x

\int \frac{x ^{2} + 2 x + 3}{( x ^{2} + 4 ) ^{2}} d x

What is the integral of e^{sin(x)+1}cos(x) ?
The integral of e^{sin(x)+1}cos(x) is e^{1+sin(x)}+C