What is the integral of 11θtan^2(θ) ?

The integral of 11θtan^2(θ) is 11(-θ^2+θtan(θ)+(θ^2)/2+ln|cos(θ)|)+C

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integral of 11θtan^2(θ)

\int 11 θ tan^{2} (θ) d θ

11 (- θ^{2} + θ tan (θ) + \frac{θ ^{2}}{2} + ln ∣ cos (θ) ∣) + C

Solution steps

\int 11 θ tan^{2} (θ) d θ

Take the constant out:

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

= 11 \cdot \int θ tan^{2} (θ) d θ

Apply Integration By Parts

: θ (- θ + tan (θ)) - \int - θ + tan (θ) d θ

= 11 (θ (- θ + tan (θ)) - \int - θ + tan (θ) d θ)

\int - θ + tan (θ) d θ = - \frac{θ ^{2}}{2} - ln ∣ cos (θ) ∣

= 11 (θ (- θ + tan (θ)) - (- \frac{θ ^{2}}{2} - ln ∣ cos (θ) ∣))

Expand

θ (- θ + tan (θ)) - (- \frac{θ ^{2}}{2} - ln ∣ cos (θ) ∣) : - θ^{2} + θ tan (θ) + \frac{θ ^{2}}{2} + ln ∣ cos (θ) ∣

= 11 (- θ^{2} + θ tan (θ) + \frac{θ ^{2}}{2} + ln ∣ cos (θ) ∣)

Add a constant to the solution

= 11 (- θ^{2} + θ tan (θ) + \frac{θ ^{2}}{2} + ln ∣ cos (θ) ∣) + C

\frac{\partial}{\partial x} (- \frac{2}{3} x - \frac{1}{3} y + 2)

x + e^{x} - \frac{1}{x}

\int e^{x} (x + 1) d x

\frac{\partial}{\partial x} (8 x e^{x y})

\frac{\partial}{\partial x} (\frac{( x - q )}{y} \cdot q)

What is the integral of 11θtan^2(θ) ?
The integral of 11θtan^2(θ) is 11(-θ^2+θtan(θ)+(θ^2)/2+ln|cos(θ)|)+C