What is the integral of (msec(θ)-tan(θ))^2 ?

The integral of (msec(θ)-tan(θ))^2 is tan(θ)-θ-2msec(θ)+m^2tan(θ)+C

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integral of (msec(θ)-tan(θ))^2

\int (m sec (θ) - tan (θ))^{2} d θ

tan (θ) - θ - 2 m sec (θ) + m^{2} tan (θ) + C

Solution steps

\int (m sec (θ) - tan (θ))^{2} d θ

Rewrite using trig identities

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

Expand

\frac{( - sin ( θ ) + m ) ^{2}}{cos ^{2} ( θ )} : \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )} - \frac{2 m sin ( θ )}{cos ^{2} ( θ )} + \frac{m ^{2}}{cos ^{2} ( θ )}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )} d θ = tan (θ) - θ

\int \frac{2 m sin ( θ )}{cos ^{2} ( θ )} d θ = 2 m sec (θ)

\int \frac{m ^{2}}{cos ^{2} ( θ )} d θ = m^{2} tan (θ)

= tan (θ) - θ - 2 m sec (θ) + m^{2} tan (θ)

Add a constant to the solution

= tan (θ) - θ - 2 m sec (θ) + m^{2} tan (θ) + C

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What is the integral of (msec(θ)-tan(θ))^2 ?
The integral of (msec(θ)-tan(θ))^2 is tan(θ)-θ-2msec(θ)+m^2tan(θ)+C