What is the integral of tsec^2(7t) ?

The integral of tsec^2(7t) is 1/2 t^2sec^2(7t)-1/98 (49t^2tan^2(7t)-14ttan(7t)-2ln|cos(7t)|+49t^2)+C

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integral of tsec^2(7t)

\int t sec^{2} (7 t) d t

\frac{1}{2} t^{2} sec^{2} (7 t) - \frac{1}{98} (49 t^{2} tan^{2} (7 t) - 14 t tan (7 t) - 2 ln ∣ cos (7 t) ∣ + 49 t^{2}) + C

Solution steps

\int t sec^{2} (7 t) d t

Rewrite using trig identities

= \int t (1 + tan^{2} (7 t)) d t

Apply Integration By Parts

: \frac{1}{2} t^{2} sec^{2} (7 t) - \int 7 t^{2} sec^{2} (7 t) tan (7 t) d t

= \frac{1}{2} t^{2} sec^{2} (7 t) - \int 7 t^{2} sec^{2} (7 t) tan (7 t) d t

\int 7 t^{2} sec^{2} (7 t) tan (7 t) d t = \frac{1}{98} (49 t^{2} tan^{2} (7 t) - 14 t tan (7 t) - 2 ln ∣ cos (7 t) ∣ + 49 t^{2})

= \frac{1}{2} t^{2} sec^{2} (7 t) - \frac{1}{98} (49 t^{2} tan^{2} (7 t) - 14 t tan (7 t) - 2 ln ∣ cos (7 t) ∣ + 49 t^{2})

Add a constant to the solution

= \frac{1}{2} t^{2} sec^{2} (7 t) - \frac{1}{98} (49 t^{2} tan^{2} (7 t) - 14 t tan (7 t) - 2 ln ∣ cos (7 t) ∣ + 49 t^{2}) + C

\int \frac{3 w - 2}{3 w ^{2} - 4 w + 1} d w

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\int x^{3.4} d x

y^{^{'}} = (\frac{x ^{2}}{y}), y (1) = 2

What is the integral of tsec^2(7t) ?
The integral of tsec^2(7t) is 1/2 t^2sec^2(7t)-1/98 (49t^2tan^2(7t)-14ttan(7t)-2ln|cos(7t)|+49t^2)+C