What is the integral of 25e^{5θ}sin(5θ) ?

The integral of 25e^{5θ}sin(5θ) is 25((e^{5θ}sin(5θ))/(10)-(e^{5θ}cos(5θ))/(10))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 25e^{5θ}sin(5θ)

\int 25 e^{5 θ} sin (5 θ) d θ

25 (\frac{e ^{5 θ} sin ( 5 θ )}{10} - \frac{e ^{5 θ} cos ( 5 θ )}{10}) + C

Solution steps

\int 25 e^{5 θ} sin (5 θ) d θ

Take the constant out:

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

= 25 \cdot \int e^{5 θ} sin (5 θ) d θ

Apply Integration By Parts

: - \frac{1}{5} e^{5 θ} cos (5 θ) - \int - e^{5 θ} cos (5 θ) d θ

= 25 (- \frac{1}{5} e^{5 θ} cos (5 θ) - \int - e^{5 θ} cos (5 θ) d θ)

Take the constant out:

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

= 25 (- \frac{1}{5} e^{5 θ} cos (5 θ) - (- \int e^{5 θ} cos (5 θ) d θ))

Apply Integration By Parts

: \frac{1}{5} e^{5 θ} sin (5 θ) - \int e^{5 θ} sin (5 θ) d θ

= 25 (- \frac{1}{5} e^{5 θ} cos (5 θ) - (- (\frac{1}{5} e^{5 θ} sin (5 θ) - \int e^{5 θ} sin (5 θ) d θ)))

Therefore

25 \cdot \int e^{5 θ} sin (5 θ) d θ = 25 (- \frac{1}{5} e^{5 θ} cos (5 θ) - (- (\frac{1}{5} e^{5 θ} sin (5 θ) - \int e^{5 θ} sin (5 θ) d θ)))

Isolate

\int e^{5 θ} sin (5 θ) d θ

= 25 (\frac{e ^{5 θ} sin ( 5 θ )}{10} - \frac{e ^{5 θ} cos ( 5 θ )}{10})

Add a constant to the solution

= 25 (\frac{e ^{5 θ} sin ( 5 θ )}{10} - \frac{e ^{5 θ} cos ( 5 θ )}{10}) + C

(2 x - 1) d x + (3 y + 4) d y = 0

\int \frac{ln ( x )}{x 1 - ln ( x )} d x

\frac{d}{d x} (15 + x^{2})

\int 17 arctan (x) d x

\frac{d}{d x} \frac{8}{( 9 x ^{2} - 7 ) ^{5}}

What is the integral of 25e^{5θ}sin(5θ) ?
The integral of 25e^{5θ}sin(5θ) is 25((e^{5θ}sin(5θ))/(10)-(e^{5θ}cos(5θ))/(10))+C