What is the integral of sec^2(x)sec^2(x) ?

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

integral of sec^2(x)sec^2(x)

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

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

Solution steps

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

Simplify

sec^{2} (x) sec^{2} (x) : sec^{4} (x)

= \int sec^{4} (x) d x

Rewrite using trig identities

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

Apply u-substitution

= \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

= \int 1 d u + \int u^{2} d u

\int 1 d u = u

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

= u + \frac{u ^{3}}{3}

Substitute back

u = tan (x)

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

Add a constant to the solution

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

\int \frac{x ^{2} + x + 1}{x ^{3} + x} d x

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\int sin (x) + sin (2 x) d x

\int \frac{( 1 + sin ( x ) )}{1 - sin ( x )} d x

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