What is the integral of 8tan^2(x) ?

The integral of 8tan^2(x) is 8(-x+tan(x))+C

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integral of 8tan^2(x)

\int 8 tan^{2} (x) d x

8 (- x + tan (x)) + C

Solution steps

\int 8 tan^{2} (x) d x

Take the constant out:

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

= 8 \cdot \int tan^{2} (x) d x

Rewrite using trig identities

= 8 \cdot \int - 1 + 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

= 8 (- \int 1 d x + \int sec^{2} (x) d x)

\int 1 d x = x

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

= 8 (- x + tan (x))

Add a constant to the solution

= 8 (- x + tan (x)) + C

\int \frac{1}{( x + 2 ) ( x + 4 )} d x

(1 + x y) d x + x^{2} d y = 0

\int - sin (x) tan (x) d x

in v erse l a pl a ce \frac{1}{( s + 2 ) ^{2} + 1}

d er i v a t i v e \frac{8 - x e ^{x}}{x + e ^{x}}