What is the integral of 28tan^6(x) ?

The integral of 28tan^6(x) is 28((tan^5(x))/5-(tan^3(x))/3+tan(x)-arctan(tan(x)))+C

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integral of 28tan^6(x)

\int 28 tan^{6} (x) d x

28 (\frac{tan ^{5} ( x )}{5} - \frac{tan ^{3} ( x )}{3} + tan (x) - arctan (tan (x))) + C

Solution steps

\int 28 tan^{6} (x) d x

Take the constant out:

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

= 28 \cdot \int tan^{6} (x) d x

Apply u-substitution

: \int \frac{u ^{6}}{1 + u ^{2}} d u

= 28 \cdot \int \frac{u ^{6}}{1 + u ^{2}} d u

Long division

\frac{u ^{6}}{1 + u ^{2}} : u^{4} - u^{2} + 1 - \frac{1}{u ^{2} + 1}

= 28 \cdot \int u^{4} - u^{2} + 1 - \frac{1}{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

= 28 (\int u^{4} d u - \int u^{2} d u + \int 1 d u - \int \frac{1}{u ^{2} + 1} d u)

\int u^{4} d u = \frac{u ^{5}}{5}

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

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

= 28 (\frac{u ^{5}}{5} - \frac{u ^{3}}{3} + u - arctan (u))

Substitute back

u = tan (x)

= 28 (\frac{tan ^{5} ( x )}{5} - \frac{tan ^{3} ( x )}{3} + tan (x) - arctan (tan (x)))

Add a constant to the solution

= 28 (\frac{tan ^{5} ( x )}{5} - \frac{tan ^{3} ( x )}{3} + tan (x) - arctan (tan (x))) + C

x \to π lim (\frac{x}{π})

\frac{d}{d x} (x + x + 5)

\int (e^{2 x} + 3)^{2} d x

\frac{152}{x}, - 49 x + 189

y^{^{''}} - 6 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = 11

What is the integral of 28tan^6(x) ?
The integral of 28tan^6(x) is 28((tan^5(x))/5-(tan^3(x))/3+tan(x)-arctan(tan(x)))+C