What is the integral of tan(x)(tan(x)+sec(x)) ?

The integral of tan(x)(tan(x)+sec(x)) is 4(-1/(2(tan(x/2)-1))-1/2 arctan(tan(x/2)))+C

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integral of tan(x)(tan(x)+sec(x))

\int tan (x) (tan (x) + sec (x)) d x

4 (- \frac{1}{2 ( tan ( \frac{x}{2} ) - 1 )} - \frac{1}{2} arctan (tan (\frac{x}{2}))) + C

Solution steps

\int tan (x) (tan (x) + sec (x)) d x

Apply u-substitution

: \int \frac{4 u}{( u - 1 ) ^{2} ( u ^{2} + 1 )} d u

= \int \frac{4 u}{( u - 1 ) ^{2} ( u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{u}{( u - 1 ) ^{2} ( u ^{2} + 1 )} d u

Take the partial fraction of

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

= 4 \cdot \int \frac{1}{2 ( u - 1 ) ^{2}} - \frac{1}{2 ( u ^{2} + 1 )} d u

Apply the Sum Rule:

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

= 4 (\int \frac{1}{2 ( u - 1 ) ^{2}} d u - \int \frac{1}{2 ( u ^{2} + 1 )} d u)

\int \frac{1}{2 ( u - 1 ) ^{2}} d u = - \frac{1}{2 ( u - 1 )}

\int \frac{1}{2 ( u ^{2} + 1 )} d u = \frac{1}{2} arctan (u)

= 4 (- \frac{1}{2 ( u - 1 )} - \frac{1}{2} arctan (u))

Substitute back

u = tan (\frac{x}{2})

= 4 (- \frac{1}{2 ( tan ( \frac{x}{2} ) - 1 )} - \frac{1}{2} arctan (tan (\frac{x}{2})))

Add a constant to the solution

= 4 (- \frac{1}{2 ( tan ( \frac{x}{2} ) - 1 )} - \frac{1}{2} arctan (tan (\frac{x}{2}))) + C

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What is the integral of tan(x)(tan(x)+sec(x)) ?
The integral of tan(x)(tan(x)+sec(x)) is 4(-1/(2(tan(x/2)-1))-1/2 arctan(tan(x/2)))+C