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

\frac{d}{d x} (- 6 sin (\frac{π}{12} x) + 18)

\int_{0}^{\frac{π}{2}} cos^{5} (y) d y

\int 8 \cdot sin^{4} (x) d x

\frac{x ^{7}}{a} + \frac{a}{b} x^{6} - c x

cos (t - 1)

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