What is the integral of tsec^2(5t) ?

The integral of tsec^2(5t) is 1/2 t^2sec^2(5t)-1/50 (25t^2tan^2(5t)-10ttan(5t)-2ln|cos(5t)|+25t^2)+C

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integral of tsec^2(5t)

\int t sec^{2} (5 t) d t

\frac{1}{2} t^{2} sec^{2} (5 t) - \frac{1}{50} (25 t^{2} tan^{2} (5 t) - 10 t tan (5 t) - 2 ln ∣ cos (5 t) ∣ + 25 t^{2}) + C

Solution steps

\int t sec^{2} (5 t) d t

Rewrite using trig identities

= \int t (1 + tan^{2} (5 t)) d t

Apply Integration By Parts

: \frac{1}{2} t^{2} sec^{2} (5 t) - \int 5 t^{2} sec^{2} (5 t) tan (5 t) d t

= \frac{1}{2} t^{2} sec^{2} (5 t) - \int 5 t^{2} sec^{2} (5 t) tan (5 t) d t

\int 5 t^{2} sec^{2} (5 t) tan (5 t) d t = \frac{1}{50} (25 t^{2} tan^{2} (5 t) - 10 t tan (5 t) - 2 ln ∣ cos (5 t) ∣ + 25 t^{2})

= \frac{1}{2} t^{2} sec^{2} (5 t) - \frac{1}{50} (25 t^{2} tan^{2} (5 t) - 10 t tan (5 t) - 2 ln ∣ cos (5 t) ∣ + 25 t^{2})

Add a constant to the solution

= \frac{1}{2} t^{2} sec^{2} (5 t) - \frac{1}{50} (25 t^{2} tan^{2} (5 t) - 10 t tan (5 t) - 2 ln ∣ cos (5 t) ∣ + 25 t^{2}) + C

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

\frac{d}{d x} (\frac{1 - ln ( x - 1 )}{( x - 1 ) ^{2}})

\frac{d}{d x} (\frac{1}{x ^{\frac{3}{2}}})

\frac{1 - 4 t}{5 + t}

\frac{3 s}{s ^{2} + 4}

What is the integral of tsec^2(5t) ?
The integral of tsec^2(5t) is 1/2 t^2sec^2(5t)-1/50 (25t^2tan^2(5t)-10ttan(5t)-2ln|cos(5t)|+25t^2)+C