What is the integral of tan^2(pi/4 x) ?

The integral of tan^2(pi/4 x) is -x+4/pi tan(pi/4 x)+C

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integral of tan^2(pi/4 x)

\int tan^{2} (\frac{π}{4} x) d x

- x + \frac{4}{π} tan (\frac{π}{4} x) + C

Solution steps

\int tan^{2} (\frac{π}{4} x) d x

Apply u-substitution

: \int tan^{2} (u) \frac{4}{π} d u

= \int tan^{2} (u) \frac{4}{π} d u

Take the constant out:

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

= \frac{4}{π} \cdot \int tan^{2} (u) d u

Rewrite using trig identities

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

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

\int 1 d u = u

\int sec^{2} (u) d u = tan (u)

= \frac{4}{π} (- u + tan (u))

Substitute back

u = \frac{π}{4} x

= \frac{4}{π} (- \frac{π}{4} x + tan (\frac{π}{4} x))

Simplify

\frac{4}{π} (- \frac{π}{4} x + tan (\frac{π}{4} x)) : - x + \frac{4}{π} tan (\frac{π}{4} x)

= - x + \frac{4}{π} tan (\frac{π}{4} x)

Add a constant to the solution

= - x + \frac{4}{π} tan (\frac{π}{4} x) + C

\frac{d Q}{d t} = k (Q - 70)

\frac{d}{d x} (- \frac{3}{2} x^{3} + \frac{2}{5} x^{2} - 4)

\int 50 x^{3} 1 - 25 x^{2} d x

\frac{d}{d x} ((3 - 5 x e^{x}) (3 x + 2))

x \to - 2 lim (\frac{- 3}{2 - x})

What is the integral of tan^2(pi/4 x) ?
The integral of tan^2(pi/4 x) is -x+4/pi tan(pi/4 x)+C