What is the integral of 3θsec^2(θ) ?

The integral of 3θsec^2(θ) is 3(θtan(θ)+ln|cos(θ)|)+C

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integral of 3θsec^2(θ)

\int 3 θ sec^{2} (θ) d θ

3 (θ tan (θ) + ln ∣ cos (θ) ∣) + C

Solution steps

\int 3 θ sec^{2} (θ) d θ

Take the constant out:

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

= 3 \cdot \int θ sec^{2} (θ) d θ

Apply Integration By Parts

: θ tan (θ) - \int tan (θ) d θ

= 3 (θ tan (θ) - \int tan (θ) d θ)

\int tan (θ) d θ = - ln ∣ cos (θ) ∣

= 3 (θ tan (θ) - (- ln ∣ cos (θ) ∣))

Simplify

= 3 (θ tan (θ) + ln ∣ cos (θ) ∣)

Add a constant to the solution

= 3 (θ tan (θ) + ln ∣ cos (θ) ∣) + C

n = 1 \sum \infty \frac{n ^{3}}{5 ^{n}}

\int \frac{7 t - 4}{t + 1} d t

\frac{d y}{d x} = 9 y \cdot e^{x + 9}

n = 1 \sum \infty \frac{n}{n ^{3} + n ^{2}}

y^{^{''}} + 2 y^{^{'}} = x^{2} + e^{- 2 x}

What is the integral of 3θsec^2(θ) ?
The integral of 3θsec^2(θ) is 3(θtan(θ)+ln|cos(θ)|)+C