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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

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

Simplify

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

= x + sin^{2} (x)

Add a constant to the solution

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

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