What is the integral of sin^2(x)e^x ?

The integral of sin^2(x)e^x is e^xsin^2(x)+2/5 e^xcos(2x)-1/5 e^xsin(2x)+C

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integral of sin^2(x)e^x

\int sin^{2} (x) e^{x} d x

e^{x} sin^{2} (x) + \frac{2}{5} e^{x} cos (2 x) - \frac{1}{5} e^{x} sin (2 x) + C

Solution steps

\int sin^{2} (x) e^{x} d x

Apply Integration By Parts

: sin^{2} (x) e^{x} - \int sin (2 x) e^{x} d x

= sin^{2} (x) e^{x} - \int sin (2 x) e^{x} d x

\int sin (2 x) e^{x} d x = - \frac{2 e ^{x} cos ( 2 x )}{5} + \frac{e ^{x} sin ( 2 x )}{5}

= sin^{2} (x) e^{x} - (- \frac{2 e ^{x} cos ( 2 x )}{5} + \frac{e ^{x} sin ( 2 x )}{5})

Simplify

sin^{2} (x) e^{x} - (- \frac{2 e ^{x} cos ( 2 x )}{5} + \frac{e ^{x} sin ( 2 x )}{5}) : e^{x} sin^{2} (x) + \frac{2}{5} e^{x} cos (2 x) - \frac{1}{5} e^{x} sin (2 x)

= e^{x} sin^{2} (x) + \frac{2}{5} e^{x} cos (2 x) - \frac{1}{5} e^{x} sin (2 x)

Add a constant to the solution

= e^{x} sin^{2} (x) + \frac{2}{5} e^{x} cos (2 x) - \frac{1}{5} e^{x} sin (2 x) + C

x^{2} sin (x)

\int x (x + 4)^{8} d x

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\int (1 - z^{- 2}) z + z^{- 1} d z

\int t sin (5 t) d t

What is the integral of sin^2(x)e^x ?
The integral of sin^2(x)e^x is e^xsin^2(x)+2/5 e^xcos(2x)-1/5 e^xsin(2x)+C