What is the integral of 16θsec^2(θ) ?

The integral of 16θsec^2(θ) is 16(θtan(θ)+ln|cos(θ)|)+C

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integral of 16θsec^2(θ)

\int 16 θ sec^{2} (θ) d θ

16 (θ tan (θ) + ln ∣ cos (θ) ∣) + C

Solution steps

\int 16 θ sec^{2} (θ) d θ

Take the constant out:

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

= 16 \cdot \int θ sec^{2} (θ) d θ

Apply Integration By Parts

: θ tan (θ) - \int tan (θ) d θ

= 16 (θ tan (θ) - \int tan (θ) d θ)

\int tan (θ) d θ = - ln ∣ cos (θ) ∣

= 16 (θ tan (θ) - (- ln ∣ cos (θ) ∣))

Simplify

= 16 (θ tan (θ) + ln ∣ cos (θ) ∣)

Add a constant to the solution

= 16 (θ tan (θ) + ln ∣ cos (θ) ∣) + C

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 4, y^{^{'}} (0) = 0

\int_{- \frac{1}{3}}^{0} \frac{3}{4} (1 - x^{2}) d x

x + x^{2} + ln (x)

x \to + 2 π lim (f (x))

ln (x^{3} + 3 x)

What is the integral of 16θsec^2(θ) ?
The integral of 16θsec^2(θ) is 16(θtan(θ)+ln|cos(θ)|)+C