What is the integral of bsin(x)cos(x) ?

The integral of bsin(x)cos(x) is b/2 sin^2(x)+C

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integral of bsin(x)cos(x)

\int b sin (x) cos (x) d x

\frac{b}{2} sin^{2} (x) + C

Solution steps

\int b sin (x) cos (x) d x

Take the constant out:

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

= b \cdot \int sin (x) cos (x) d x

Apply u-substitution

: \int u d u

= b \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= b \frac{u ^{2}}{2}

Substitute back

u = sin (x)

= b \frac{sin ^{2} ( x )}{2}

Simplify

b \frac{sin ^{2} ( x )}{2} : \frac{b}{2} sin^{2} (x)

= \frac{b}{2} sin^{2} (x)

Add a constant to the solution

= \frac{b}{2} sin^{2} (x) + C

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

\int \frac{1}{3 y - 1} d y

\int (tan^{2} (3 x)) + 10 d x

\int \frac{sin ( 4 t - 1 )}{1 - sin ^{2} ( 4 t - 1 )} d t

\int (a + b x + x^{5}) d x

What is the integral of bsin(x)cos(x) ?
The integral of bsin(x)cos(x) is b/2 sin^2(x)+C