What is the integral of tsec^2(6t) ?

The integral of tsec^2(6t) is 1/2 t^2sec^2(6t)-1/36 (18t^2tan^2(6t)-6ttan(6t)-ln|cos(6t)|+18t^2)+C

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integral of tsec^2(6t)

\int t sec^{2} (6 t) d t

\frac{1}{2} t^{2} sec^{2} (6 t) - \frac{1}{36} (18 t^{2} tan^{2} (6 t) - 6 t tan (6 t) - ln ∣ cos (6 t) ∣ + 18 t^{2}) + C

Solution steps

\int t sec^{2} (6 t) d t

Rewrite using trig identities

= \int t (1 + tan^{2} (6 t)) d t

Apply Integration By Parts

: \frac{1}{2} t^{2} sec^{2} (6 t) - \int 6 t^{2} sec^{2} (6 t) tan (6 t) d t

= \frac{1}{2} t^{2} sec^{2} (6 t) - \int 6 t^{2} sec^{2} (6 t) tan (6 t) d t

\int 6 t^{2} sec^{2} (6 t) tan (6 t) d t = \frac{1}{36} (18 t^{2} tan^{2} (6 t) - 6 t tan (6 t) - ln ∣ cos (6 t) ∣ + 18 t^{2})

= \frac{1}{2} t^{2} sec^{2} (6 t) - \frac{1}{36} (18 t^{2} tan^{2} (6 t) - 6 t tan (6 t) - ln ∣ cos (6 t) ∣ + 18 t^{2})

Add a constant to the solution

= \frac{1}{2} t^{2} sec^{2} (6 t) - \frac{1}{36} (18 t^{2} tan^{2} (6 t) - 6 t tan (6 t) - ln ∣ cos (6 t) ∣ + 18 t^{2}) + C

n = 0 \sum \infty \frac{1}{1. 1 ^{n}}

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x y^{^{'}} + 2 y = - 3 sin (x)

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\int \frac{1}{x ln ( x ^{5} )} d x

What is the integral of tsec^2(6t) ?
The integral of tsec^2(6t) is 1/2 t^2sec^2(6t)-1/36 (18t^2tan^2(6t)-6ttan(6t)-ln|cos(6t)|+18t^2)+C