What is the integral of tan^6(x)sec^2(x) ?

The integral of tan^6(x)sec^2(x) is (tan^7(x))/7+C

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integral of tan^6(x)sec^2(x)

\int tan^{6} (x) sec^{2} (x) d x

\frac{tan ^{7} ( x )}{7} + C

Solution steps

\int tan^{6} (x) sec^{2} (x) d x

Rewrite using trig identities

= \int (1 + tan^{2} (x)) tan^{6} (x) d x

Apply u-substitution

: \int u^{6} d u

= \int u^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= \frac{u ^{7}}{7}

Substitute back

u = tan (x)

= \frac{tan ^{7} ( x )}{7}

Add a constant to the solution

= \frac{tan ^{7} ( x )}{7} + C

x^{4} + 6 x

f (x) = e^{x} sin (x)

\frac{d}{d x} (x^{2} x^{2} + 1)

x \to 0 - lim (\frac{5 x}{x})

\int y sin (y) d y

What is the integral of tan^6(x)sec^2(x) ?
The integral of tan^6(x)sec^2(x) is (tan^7(x))/7+C