What is the integral of 9tsec^2(2t) ?

The integral of 9tsec^2(2t) is 9(1/2 ttan(2t)+1/4 ln|cos(2t)|)+C

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integral of 9tsec^2(2t)

\int 9 t sec^{2} (2 t) d t

9 (\frac{1}{2} t tan (2 t) + \frac{1}{4} ln ∣ cos (2 t) ∣) + C

Solution steps

\int 9 t sec^{2} (2 t) d t

Take the constant out:

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

= 9 \cdot \int t sec^{2} (2 t) d t

Apply Integration By Parts

: \frac{1}{2} t tan (2 t) - \int \frac{1}{2} tan (2 t) d t

= 9 (\frac{1}{2} t tan (2 t) - \int \frac{1}{2} tan (2 t) d t)

\int \frac{1}{2} tan (2 t) d t = - \frac{1}{4} ln ∣ cos (2 t) ∣

= 9 (\frac{1}{2} t tan (2 t) - (- \frac{1}{4} ln ∣ cos (2 t) ∣))

Simplify

= 9 (\frac{1}{2} t tan (2 t) + \frac{1}{4} ln ∣ cos (2 t) ∣)

Add a constant to the solution

= 9 (\frac{1}{2} t tan (2 t) + \frac{1}{4} ln ∣ cos (2 t) ∣) + C

\int 7 sin^{2} (x) cos^{3} (x) d x

n = 0 \sum \infty \frac{8 ^{n}}{n !}

e^{2 t} sin (t)

\frac{d y}{d x}, y = \frac{3}{x}

x - y cos (\frac{y}{x}) d x + x cos (\frac{y}{x}) d y = 0

What is the integral of 9tsec^2(2t) ?
The integral of 9tsec^2(2t) is 9(1/2 ttan(2t)+1/4 ln|cos(2t)|)+C