What is the integral of (x^4sin(y)-y^2cos(x)+8^x) ?

The integral of (x^4sin(y)-y^2cos(x)+8^x) is (8^x)/(3ln(2))-y^2sin(x)+1/5 x^5sin(y)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^4sin(y)-y^2cos(x)+8^x)

\int (x^{4} sin (y) - y^{2} cos (x) + 8^{x}) d x

\frac{8 ^{x}}{3 ln ( 2 )} - y^{2} sin (x) + \frac{1}{5} x^{5} sin (y) + C

Solution steps

\int x^{4} sin (y) - y^{2} cos (x) + 8^{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 8^{x} d x - \int cos (x) y^{2} d x + \int sin (y) x^{4} d x

\int 8^{x} d x = \frac{8 ^{x}}{3 ln ( 2 )}

\int cos (x) y^{2} d x = y^{2} sin (x)

\int sin (y) x^{4} d x = \frac{1}{5} x^{5} sin (y)

= \frac{8 ^{x}}{3 ln ( 2 )} - y^{2} sin (x) + \frac{1}{5} x^{5} sin (y)

Add a constant to the solution

= \frac{8 ^{x}}{3 ln ( 2 )} - y^{2} sin (x) + \frac{1}{5} x^{5} sin (y) + C

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

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

\int x e^{- 3 x^{2} + 9} d x

\int \frac{x - 6}{x ^{2} ( x + 2 )} d x

\int t^{2} sin (Bt) d t

What is the integral of (x^4sin(y)-y^2cos(x)+8^x) ?
The integral of (x^4sin(y)-y^2cos(x)+8^x) is (8^x)/(3ln(2))-y^2sin(x)+1/5 x^5sin(y)+C