What is the integral of 1/(cos^5(x)) ?

The integral of 1/(cos^5(x)) is (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 1/(cos^5(x))

\int \frac{1}{cos ^{5} ( x )} d x

\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 \frac{1}{cos ^{5} ( x )} d x

Rewrite using trig identities

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

Apply Integral Reduction

= \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) ∣

= \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

= \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

2 y y^{^{'}} + 2 = y^{2} + 2 x, y (0) = 6

\frac{2}{x ^{\frac{1}{3}}} - 2

2 y^{^{'}} + t y = 7

\frac{d}{d x} (\frac{3 x - 1}{2 x + 5})

y = 4 x^{2}, y^{2} = \frac{1}{4} x

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