What is the integral of 10sec(2x)tan(2x) ?

The integral of 10sec(2x)tan(2x) is 10(1/2 cos(2x)+1/2 sin(2x)tan(2x))+C

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integral of 10sec(2x)tan(2x)

\int 10 sec (2 x) tan (2 x) d x

10 (\frac{1}{2} cos (2 x) + \frac{1}{2} sin (2 x) tan (2 x)) + C

Solution steps

\int 10 sec (2 x) tan (2 x) d x

Take the constant out:

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

= 10 \cdot \int sec (2 x) tan (2 x) d x

Rewrite using trig identities

= 10 \cdot \int \frac{sin ( 2 x )}{cos ^{2} ( 2 x )} d x

Apply Integration By Parts

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

= 10 (- \int sin (2 x) d x + \frac{1}{2} sin (2 x) tan (2 x))

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

= 10 (- (- \frac{1}{2} cos (2 x)) + \frac{1}{2} sin (2 x) tan (2 x))

Simplify

= 10 (\frac{1}{2} cos (2 x) + \frac{1}{2} sin (2 x) tan (2 x))

Add a constant to the solution

= 10 (\frac{1}{2} cos (2 x) + \frac{1}{2} sin (2 x) tan (2 x)) + C

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\frac{\partial}{\partial y} (x^{2} y^{2} - x)

3 + t^{5} + sin (π t)

What is the integral of 10sec(2x)tan(2x) ?
The integral of 10sec(2x)tan(2x) is 10(1/2 cos(2x)+1/2 sin(2x)tan(2x))+C