What is the integral of θcos^2(θ) ?

The integral of θcos^2(θ) is 1/2 θ(θ+1/2 sin(2θ))-1/2 ((θ^2)/2-1/4 cos(2θ))+C

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integral of θcos^2(θ)

\int θ cos^{2} (θ) d θ

\frac{1}{2} θ (θ + \frac{1}{2} sin (2 θ)) - \frac{1}{2} (\frac{θ ^{2}}{2} - \frac{1}{4} cos (2 θ)) + C

Solution steps

\int θ cos^{2} (θ) d θ

Apply Integration By Parts

: \frac{1}{2} θ (θ + \frac{1}{2} sin (2 θ)) - \int \frac{1}{2} (θ + \frac{1}{2} sin (2 θ)) d θ

= \frac{1}{2} θ (θ + \frac{1}{2} sin (2 θ)) - \int \frac{1}{2} (θ + \frac{1}{2} sin (2 θ)) d θ

\int \frac{1}{2} (θ + \frac{1}{2} sin (2 θ)) d θ = \frac{1}{2} (\frac{θ ^{2}}{2} - \frac{1}{4} cos (2 θ))

= \frac{1}{2} θ (θ + \frac{1}{2} sin (2 θ)) - \frac{1}{2} (\frac{θ ^{2}}{2} - \frac{1}{4} cos (2 θ))

Add a constant to the solution

= \frac{1}{2} θ (θ + \frac{1}{2} sin (2 θ)) - \frac{1}{2} (\frac{θ ^{2}}{2} - \frac{1}{4} cos (2 θ)) + C

y = (\frac{x}{7} + 9)^{\frac{3}{2}}

\frac{d}{d x} (\frac{x - 1}{x ^{2} + x - 2})

\frac{d y}{d x} = x + x y

f (x) = 2 ln (x)

\frac{- s - 1}{s ^{2} + s + 1}

What is the integral of θcos^2(θ) ?
The integral of θcos^2(θ) is 1/2 θ(θ+1/2 sin(2θ))-1/2 ((θ^2)/2-1/4 cos(2θ))+C