What is the integral of 8tan^7(x)sec^2(x) ?

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

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integral of 8tan^7(x)sec^2(x)

\int 8 tan^{7} (x) sec^{2} (x) d x

tan^{8} (x) + C

Solution steps

\int 8 tan^{7} (x) sec^{2} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int tan^{7} (x) sec^{2} (x) d x

Rewrite using trig identities

= 8 \cdot \int (1 + tan^{2} (x)) tan^{7} (x) d x

Apply u-substitution

: \int u^{7} d u

= 8 \cdot \int u^{7} d u

Apply the Power Rule

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

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

Substitute back

u = tan (x)

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

Simplify

8 \cdot \frac{tan ^{8} ( x )}{8} : tan^{8} (x)

= tan^{8} (x)

Add a constant to the solution

= tan^{8} (x) + C

\frac{d y}{d x} = 8 - y

\frac{d}{d x} (\frac{x ^{3}}{3} - 4 x)

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

2 x^{3} - e^{x y} = x \cdot y^{4}

x \to 0 + lim (x^{9 x})

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