What is the integral of 5^{sin(2x)}cos(2x) ?

The integral of 5^{sin(2x)}cos(2x) is (5^{sin(2x)})/(2ln(5))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 5^{sin(2x)}cos(2x)

\int 5^{s i n (2 x)} cos (2 x) d x

\frac{5 ^{s i n (2 x)}}{2 ln ( 5 )} + C

Solution steps

\int 5^{s i n (2 x)} cos (2 x) d x

Apply u-substitution

: \int 5^{s i n (u)} cos (u) \frac{1}{2} d u

= \int 5^{s i n (u)} cos (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int 5^{s i n (u)} cos (u) d u

Apply u-substitution

: \int 5^{v} d v

= \frac{1}{2} \cdot \int 5^{v} d v

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

= \frac{1}{2} \cdot \frac{5 ^{v}}{ln ( 5 )}

Substitute back

= \frac{1}{2} \cdot \frac{5 ^{s i n (2 x)}}{ln ( 5 )}

Simplify

\frac{1}{2} \cdot \frac{5 ^{s i n (2 x)}}{ln ( 5 )} : \frac{5 ^{s i n (2 x)}}{2 ln ( 5 )}

= \frac{5 ^{s i n (2 x)}}{2 ln ( 5 )}

Add a constant to the solution

= \frac{5 ^{s i n (2 x)}}{2 ln ( 5 )} + C

x \to 0 lim (6 x^{3} + 7)

n = 1 \sum \infty \frac{1}{n}

\int 1 + y e^{x y} d x

\int - 8 (x - 1) \cdot (1 - x) d x

t^{2} y^{^{''}} + t y^{^{'}} + 9 y = 0

What is the integral of 5^{sin(2x)}cos(2x) ?
The integral of 5^{sin(2x)}cos(2x) is (5^{sin(2x)})/(2ln(5))+C