What is the integral of 29tan^6(x) ?

The integral of 29tan^6(x) is 29((tan^5(x))/5-(tan^3(x))/3+tan(x)-arctan(tan(x)))+C

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integral of 29tan^6(x)

\int 29 tan^{6} (x) d x

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

Solution steps

\int 29 tan^{6} (x) d x

Take the constant out:

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

= 29 \cdot \int tan^{6} (x) d x

Apply u-substitution

= 29 \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}

= 29 \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

= 29 (\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)

= 29 (\frac{u ^{5}}{5} - \frac{u ^{3}}{3} + u - arctan (u))

Substitute back

u = tan (x)

= 29 (\frac{tan ^{5} ( x )}{5} - \frac{tan ^{3} ( x )}{3} + tan (x) - arctan (tan (x)))

Add a constant to the solution

= 29 (\frac{tan ^{5} ( x )}{5} - \frac{tan ^{3} ( x )}{3} + tan (x) - arctan (tan (x))) + C

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

\int \frac{20}{20 + e ^{x}} d x

\int 12 x^{3} (3 x^{4} + 4)^{4} d x

\int \frac{1}{u ^{2} - 9} d u

\int (t) d t

What is the integral of 29tan^6(x) ?
The integral of 29tan^6(x) is 29((tan^5(x))/5-(tan^3(x))/3+tan(x)-arctan(tan(x)))+C