What is the integral of-5tan^5(4x)sec^6(4x) ?

The integral of-5tan^5(4x)sec^6(4x) is -5((sec^{10}(4x))/(40)-(sec^8(4x))/(16)+(sec^6(4x))/(24))+C

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integral of-5tan^5(4x)sec^6(4x)

\int - 5 tan^{5} (4 x) sec^{6} (4 x) d x

- 5 (\frac{sec ^{10} ( 4 x )}{40} - \frac{sec ^{8} ( 4 x )}{16} + \frac{sec ^{6} ( 4 x )}{24}) + C

Solution steps

\int - 5 tan^{5} (4 x) sec^{6} (4 x) d x

Take the constant out:

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

= - 5 \cdot \int tan^{5} (4 x) sec^{6} (4 x) d x

Rewrite using trig identities

= - 5 \cdot \int (- 1 + sec^{2} (4 x))^{2} tan (4 x) sec^{6} (4 x) d x

Apply u-substitution

: \int \frac{u ^{5} ( u ^{2} - 1 ) ^{2}}{4} d u

= - 5 \cdot \int \frac{u ^{5} ( u ^{2} - 1 ) ^{2}}{4} d u

Expand

\frac{u ^{5} ( u ^{2} - 1 ) ^{2}}{4} : \frac{u ^{9}}{4} - \frac{u ^{7}}{2} + \frac{u ^{5}}{4}

= - 5 \cdot \int \frac{u ^{9}}{4} - \frac{u ^{7}}{2} + \frac{u ^{5}}{4} 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

= - 5 (\int \frac{u ^{9}}{4} d u - \int \frac{u ^{7}}{2} d u + \int \frac{u ^{5}}{4} d u)

\int \frac{u ^{9}}{4} d u = \frac{u ^{10}}{40}

\int \frac{u ^{7}}{2} d u = \frac{u ^{8}}{16}

\int \frac{u ^{5}}{4} d u = \frac{u ^{6}}{24}

= - 5 (\frac{u ^{10}}{40} - \frac{u ^{8}}{16} + \frac{u ^{6}}{24})

Substitute back

u = sec (4 x)

= - 5 (\frac{sec ^{10} ( 4 x )}{40} - \frac{sec ^{8} ( 4 x )}{16} + \frac{sec ^{6} ( 4 x )}{24})

Add a constant to the solution

= - 5 (\frac{sec ^{10} ( 4 x )}{40} - \frac{sec ^{8} ( 4 x )}{16} + \frac{sec ^{6} ( 4 x )}{24}) + C

\int_{0}^{1} \frac{1}{x ^{0.1}} d x

x \to 0 lim (\frac{tan ( 9 x )}{sin ( 4 x )})

\frac{\partial}{\partial x} (x e^{y})

((e^{2 x} - 5 x)^{2})^{^{'}}

What is the integral of-5tan^5(4x)sec^6(4x) ?
The integral of-5tan^5(4x)sec^6(4x) is -5((sec^{10}(4x))/(40)-(sec^8(4x))/(16)+(sec^6(4x))/(24))+C