What is the integral of 8cos(θ)sin(θ) ?

The integral of 8cos(θ)sin(θ) is 4sin^2(θ)+C

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integral of 8cos(θ)sin(θ)

\int 8 cos (θ) sin (θ) d θ

4 sin^{2} (θ) + C

Solution steps

\int 8 cos (θ) sin (θ) d θ

Take the constant out:

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

= 8 \cdot \int cos (θ) sin (θ) d θ

Apply u-substitution

: \int u d u

= 8 \cdot \int u d u

Apply the Power Rule

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

= 8 \cdot \frac{u ^{2}}{2}

Substitute back

u = sin (θ)

= 8 \cdot \frac{sin ^{2} ( θ )}{2}

Simplify

8 \cdot \frac{sin ^{2} ( θ )}{2} : 4 sin^{2} (θ)

= 4 sin^{2} (θ)

Add a constant to the solution

= 4 sin^{2} (θ) + C

\frac{\partial}{\partial u} (3 cos (u))

\int_{1}^{2} w^{7} e^{w^{4}} d w

y^{^{'''}} + y^{^{''}} - y^{^{'}} - y = 0

\frac{d}{d x} (\frac{36}{( 1 + 3 x ) ^{3}})

(2, 2), (19, - 17)

What is the integral of 8cos(θ)sin(θ) ?
The integral of 8cos(θ)sin(θ) is 4sin^2(θ)+C