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

The integral of ((1+tan(t))^2)/(sec^2(t)) is t+sin^2(t)+C

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

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

t + sin^{2} (t) + C

Solution steps

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

Expand

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

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

\int \frac{1}{sec ^{2} ( t )} d t = \frac{1}{2} (t + \frac{1}{2} sin (2 t))

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

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

= \frac{1}{2} (t + \frac{1}{2} sin (2 t)) + sin^{2} (t) + \frac{1}{2} (t - \frac{1}{2} sin (2 t))

Simplify

\frac{1}{2} (t + \frac{1}{2} sin (2 t)) + sin^{2} (t) + \frac{1}{2} (t - \frac{1}{2} sin (2 t)) : t + sin^{2} (t)

= t + sin^{2} (t)

Add a constant to the solution

= t + sin^{2} (t) + C

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