What is the integral of (sec^4(x)-2sec^2(x)+1) ?

The integral of (sec^4(x)-2sec^2(x)+1) is x-tan(x)+1/3 tan^3(x)+C

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integral of (sec^4(x)-2sec^2(x)+1)

\int (sec^{4} (x) - 2 sec^{2} (x) + 1) d x

x - tan (x) + \frac{1}{3} tan^{3} (x) + C

Solution steps

\int sec^{4} (x) - 2 sec^{2} (x) + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int sec^{4} (x) d x = tan (x) + \frac{tan ^{3} ( x )}{3}

\int 2 sec^{2} (x) d x = 2 tan (x)

\int 1 d x = x

= tan (x) + \frac{tan ^{3} ( x )}{3} - 2 tan (x) + x

Simplify

tan (x) + \frac{tan ^{3} ( x )}{3} - 2 tan (x) + x : x - tan (x) + \frac{1}{3} tan^{3} (x)

= x - tan (x) + \frac{1}{3} tan^{3} (x)

Add a constant to the solution

= x - tan (x) + \frac{1}{3} tan^{3} (x) + C

\int + \frac{3 x ^{2} - 21 x + 32}{x ^{3} - 8 x ^{2} + 16 x} d x

\frac{x}{x ^{2} + 2}

\int \frac{1}{t - 1 - t ^{2}} d t

\int 21 sin^{4} (x) d x

\int \frac{2 u}{1 - u} d u

What is the integral of (sec^4(x)-2sec^2(x)+1) ?
The integral of (sec^4(x)-2sec^2(x)+1) is x-tan(x)+1/3 tan^3(x)+C