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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

= \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

\frac{\partial}{\partial y} (10 e^{x} sin (y) z^{3})

f (θ) = 3 cos (θ)

\int \frac{1}{cos ^{2} ( t ) 1 + tan ( t )} d t

x \to 2 lim (\frac{5}{( x - 5 ) ^{2}})

3 y^{^{''}} - 6 y^{^{'}} + 6 y = 0

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