What is the integral of tan^7(x/4)sec^2(x/4) ?

The integral of tan^7(x/4)sec^2(x/4) is 1/2 tan^8(x/4)+C

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integral of tan^7(x/4)sec^2(x/4)

\int tan^{7} (\frac{x}{4}) sec^{2} (\frac{x}{4}) d x

\frac{1}{2} tan^{8} (\frac{x}{4}) + C

Solution steps

\int tan^{7} (\frac{x}{4}) sec^{2} (\frac{x}{4}) d x

Apply u-substitution

= \int tan^{7} (u) sec^{2} (u) \cdot 4 d u

Take the constant out:

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

= 4 \cdot \int tan^{7} (u) sec^{2} (u) d u

Rewrite using trig identities

= 4 \cdot \int (1 + tan^{2} (u)) tan^{7} (u) d u

Apply u-substitution

= 4 \cdot \int v^{7} d v

Apply the Power Rule

= 4 \cdot \frac{v ^{8}}{8}

Substitute back

= 4 \cdot \frac{tan ^{8} ( \frac{x}{4} )}{8}

Simplify

4 \cdot \frac{tan ^{8} ( \frac{x}{4} )}{8} : \frac{1}{2} tan^{8} (\frac{x}{4})

= \frac{1}{2} tan^{8} (\frac{x}{4})

Add a constant to the solution

= \frac{1}{2} tan^{8} (\frac{x}{4}) + C

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What is the integral of tan^7(x/4)sec^2(x/4) ?
The integral of tan^7(x/4)sec^2(x/4) is 1/2 tan^8(x/4)+C