What is the integral of 11tan^3(x)sec(x) ?

The integral of 11tan^3(x)sec(x) is 11(-sec(x)+(sec^3(x))/3)+C

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integral of 11tan^3(x)sec(x)

\int 11 tan^{3} (x) sec (x) d x

11 (- sec (x) + \frac{sec ^{3} ( x )}{3}) + C

Solution steps

\int 11 tan^{3} (x) sec (x) d x

Take the constant out:

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

= 11 \cdot \int tan^{3} (x) sec (x) d x

Rewrite using trig identities

= 11 \cdot \int (- 1 + sec^{2} (x)) tan (x) sec (x) d x

Apply u-substitution

: \int - 1 + u^{2} d u

= 11 \cdot \int - 1 + u^{2} 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

= 11 (- \int 1 d u + \int u^{2} d u)

\int 1 d u = u

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

= 11 (- u + \frac{u ^{3}}{3})

Substitute back

u = sec (x)

= 11 (- sec (x) + \frac{sec ^{3} ( x )}{3})

Add a constant to the solution

= 11 (- sec (x) + \frac{sec ^{3} ( x )}{3}) + C

\frac{d}{d x} ((sin (5 x))^{l n (x)})

\frac{d}{d x} (\frac{81}{x ^{4}})

y^{^{'}} - \frac{y}{x} = tan (\frac{y}{x})

\int \frac{x}{( x ^{2} + 4 x + 5 ) ^{\frac{3}{2}}} d x

\frac{d y}{d x} = \frac{( x - 4 )}{( y + 5 )}

What is the integral of 11tan^3(x)sec(x) ?
The integral of 11tan^3(x)sec(x) is 11(-sec(x)+(sec^3(x))/3)+C