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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

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

Substitute back

u = tan (x)

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

Add a constant to the solution

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

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

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\int \frac{1}{x ^{2} + 6 x + 8} d x

\int \frac{x ^{3} - 3 x}{x ^{4} - 6 x ^{2} - 9} d x

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