What is the integral of (1+tan(3t))^3sec^2(3t) ?

The integral of (1+tan(3t))^3sec^2(3t) is 1/3 tan(3t)+1/2 tan^2(3t)+1/3 tan^3(3t)+1/12 tan^4(3t)+C

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integral of (1+tan(3t))^3sec^2(3t)

\int (1 + tan (3 t))^{3} sec^{2} (3 t) d t

\frac{1}{3} tan (3 t) + \frac{1}{2} tan^{2} (3 t) + \frac{1}{3} tan^{3} (3 t) + \frac{1}{12} tan^{4} (3 t) + C

Solution steps

\int (1 + tan (3 t))^{3} sec^{2} (3 t) d t

Expand

(1 + tan (3 t))^{3} sec^{2} (3 t) : sec^{2} (3 t) + 3 sec^{2} (3 t) tan (3 t) + 3 sec^{2} (3 t) tan^{2} (3 t) + tan^{3} (3 t) sec^{2} (3 t)

= \int sec^{2} (3 t) + 3 sec^{2} (3 t) tan (3 t) + 3 sec^{2} (3 t) tan^{2} (3 t) + tan^{3} (3 t) sec^{2} (3 t) d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int sec^{2} (3 t) d t + \int 3 sec^{2} (3 t) tan (3 t) d t + \int 3 sec^{2} (3 t) tan^{2} (3 t) d t + \int tan^{3} (3 t) sec^{2} (3 t) d t

\int sec^{2} (3 t) d t = \frac{1}{3} tan (3 t)

\int 3 sec^{2} (3 t) tan (3 t) d t = \frac{1}{2} tan^{2} (3 t)

\int 3 sec^{2} (3 t) tan^{2} (3 t) d t = \frac{1}{3} tan^{3} (3 t)

\int tan^{3} (3 t) sec^{2} (3 t) d t = \frac{1}{12} tan^{4} (3 t)

= \frac{1}{3} tan (3 t) + \frac{1}{2} tan^{2} (3 t) + \frac{1}{3} tan^{3} (3 t) + \frac{1}{12} tan^{4} (3 t)

Add a constant to the solution

= \frac{1}{3} tan (3 t) + \frac{1}{2} tan^{2} (3 t) + \frac{1}{3} tan^{3} (3 t) + \frac{1}{12} tan^{4} (3 t) + C

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What is the integral of (1+tan(3t))^3sec^2(3t) ?
The integral of (1+tan(3t))^3sec^2(3t) is 1/3 tan(3t)+1/2 tan^2(3t)+1/3 tan^3(3t)+1/12 tan^4(3t)+C