What is the integral of 29tan^5(x)sec^4(x) ?

The integral of 29tan^5(x)sec^4(x) is 29((tan^6(x))/6+(tan^8(x))/8)+C

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integral of 29tan^5(x)sec^4(x)

\int 29 tan^{5} (x) sec^{4} (x) d x

29 (\frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{8}) + C

Solution steps

\int 29 tan^{5} (x) sec^{4} (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^{5} (x) sec^{4} (x) d x

Rewrite using trig identities

= 29 \cdot \int (1 + tan^{2} (x)) sec^{2} (x) tan^{5} (x) d x

Apply u-substitution

: \int u^{5} (1 + u^{2}) d u

= 29 \cdot \int u^{5} (1 + u^{2}) d u

Expand

u^{5} (1 + u^{2}) : u^{5} + u^{7}

= 29 \cdot \int u^{5} + u^{7} 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^{5} d u + \int u^{7} d u)

\int u^{5} d u = \frac{u ^{6}}{6}

\int u^{7} d u = \frac{u ^{8}}{8}

= 29 (\frac{u ^{6}}{6} + \frac{u ^{8}}{8})

Substitute back

u = tan (x)

= 29 (\frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{8})

Add a constant to the solution

= 29 (\frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{8}) + C

\int_{0}^{4} 7 e^{- x^{2}} d x

F (x) = (7 x^{6} + 4 x^{3})^{4}

x \to 0 lim (\frac{1}{x - 4})

\int sin (x) sin (3 x) d x

\frac{d y}{d x}, y = x^{3 x}

What is the integral of 29tan^5(x)sec^4(x) ?
The integral of 29tan^5(x)sec^4(x) is 29((tan^6(x))/6+(tan^8(x))/8)+C