What is the integral of tan^7(x)sec^2(x) ?

The integral of tan^7(x)sec^2(x) is (tan^8(x))/8+C

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integral of tan^7(x)sec^2(x)

\int tan^{7} (x) sec^{2} (x) d x

\frac{tan ^{8} ( x )}{8} + C

Solution steps

\int tan^{7} (x) sec^{2} (x) d x

Rewrite using trig identities

= \int (1 + tan^{2} (x)) tan^{7} (x) d x

Apply u-substitution

: \int u^{7} d u

= \int u^{7} d u

Apply the Power Rule

: \frac{u ^{8}}{8}

= \frac{u ^{8}}{8}

Substitute back

u = tan (x)

= \frac{tan ^{8} ( x )}{8}

Add a constant to the solution

= \frac{tan ^{8} ( x )}{8} + C

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

x \to 4 + lim (ln (x^{2} - 16))

\frac{1}{2} y^{^{'}} + y = 0, y (1) = 1

f (x) = 1 - x^{2}, (1, 0)

\frac{d}{d x} (\frac{( 6 x ^{2} + 6 x + 8 )}{x})

What is the integral of tan^7(x)sec^2(x) ?
The integral of tan^7(x)sec^2(x) is (tan^8(x))/8+C