What is the integral of 24sec^5(x) ?

The integral of 24sec^5(x) is 24((sec^4(x)sin(x))/4+3/4 (1/2 sec(x)tan(x)+1/2 ln|tan(x)+sec(x)|))+C

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integral of 24sec^5(x)

\int 24 sec^{5} (x) d x

24 (\frac{sec ^{4} ( x ) sin ( x )}{4} + \frac{3}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)) + C

Solution steps

\int 24 sec^{5} (x) d x

Take the constant out:

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

= 24 \cdot \int sec^{5} (x) d x

Apply Integral Reduction

= 24 (\frac{sec ^{4} ( x ) sin ( x )}{4} + \frac{3}{4} \cdot \int sec^{3} (x) d x)

\int sec^{3} (x) d x = \frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣

= 24 (\frac{sec ^{4} ( x ) sin ( x )}{4} + \frac{3}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣))

Add a constant to the solution

= 24 (\frac{sec ^{4} ( x ) sin ( x )}{4} + \frac{3}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)) + C

\frac{d y}{d x} = \frac{( y ^{2} - 1 )}{( x ^{2} - 1 )}, y (0) = 1

\frac{\partial}{\partial x} (4 x cos (x y))

f (x) = x^{2} + x + 2, 0, 12

\int \frac{4 + 7 x}{1 + x ^{2}} d x

\frac{d}{d x} (6 x (x^{2} - 1)^{2})

What is the integral of 24sec^5(x) ?
The integral of 24sec^5(x) is 24((sec^4(x)sin(x))/4+3/4 (1/2 sec(x)tan(x)+1/2 ln|tan(x)+sec(x)|))+C