What is the integral of pi/2 pi/(32)sec(2θ)tan(2θ) ?

The integral of pi/2 pi/(32)sec(2θ)tan(2θ) is (pi^2)/(64)(1/2 cos(2θ)+1/2 sin(2θ)tan(2θ))+C

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integral of pi/2 pi/(32)sec(2θ)tan(2θ)

\int \frac{π}{2} \frac{π}{32} sec (2 θ) tan (2 θ) d θ

\frac{π ^{2}}{64} (\frac{1}{2} cos (2 θ) + \frac{1}{2} sin (2 θ) tan (2 θ)) + C

Solution steps

\int \frac{π}{2} \cdot \frac{π}{32} sec (2 θ) tan (2 θ) d θ

Take the constant out:

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

= \frac{π}{2} \cdot \frac{π}{32} \cdot \int sec (2 θ) tan (2 θ) d θ

\frac{π}{2} \cdot \frac{π}{32} = \frac{π ^{2}}{64}

= \frac{π ^{2}}{64} \cdot \int sec (2 θ) tan (2 θ) d θ

Rewrite using trig identities

= \frac{π ^{2}}{64} \cdot \int \frac{sin ( 2 θ )}{cos ^{2} ( 2 θ )} d θ

Apply Integration By Parts

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

= \frac{π ^{2}}{64} (- \int sin (2 θ) d θ + \frac{1}{2} sin (2 θ) tan (2 θ))

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

= \frac{π ^{2}}{64} (- (- \frac{1}{2} cos (2 θ)) + \frac{1}{2} sin (2 θ) tan (2 θ))

Simplify

= \frac{π ^{2}}{64} (\frac{1}{2} cos (2 θ) + \frac{1}{2} sin (2 θ) tan (2 θ))

Add a constant to the solution

= \frac{π ^{2}}{64} (\frac{1}{2} cos (2 θ) + \frac{1}{2} sin (2 θ) tan (2 θ)) + C

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What is the integral of pi/2 pi/(32)sec(2θ)tan(2θ) ?
The integral of pi/2 pi/(32)sec(2θ)tan(2θ) is (pi^2)/(64)(1/2 cos(2θ)+1/2 sin(2θ)tan(2θ))+C