What is the integral of 63sec^4(θ)tan^6(θ) ?

The integral of 63sec^4(θ)tan^6(θ) is 63((tan^7(θ))/7+(tan^9(θ))/9)+C

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integral of 63sec^4(θ)tan^6(θ)

\int 63 sec^{4} (θ) tan^{6} (θ) d θ

63 (\frac{tan ^{7} ( θ )}{7} + \frac{tan ^{9} ( θ )}{9}) + C

Solution steps

\int 63 sec^{4} (θ) tan^{6} (θ) d θ

Take the constant out:

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

= 63 \cdot \int sec^{4} (θ) tan^{6} (θ) d θ

Rewrite using trig identities

= 63 \cdot \int (1 + tan^{2} (θ)) sec^{2} (θ) tan^{6} (θ) d θ

Apply u-substitution

: \int u^{6} (1 + u^{2}) d u

= 63 \cdot \int u^{6} (1 + u^{2}) d u

Expand

u^{6} (1 + u^{2}) : u^{6} + u^{8}

= 63 \cdot \int u^{6} + u^{8} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 63 (\int u^{6} d u + \int u^{8} d u)

\int u^{6} d u = \frac{u ^{7}}{7}

\int u^{8} d u = \frac{u ^{9}}{9}

= 63 (\frac{u ^{7}}{7} + \frac{u ^{9}}{9})

Substitute back

u = tan (θ)

= 63 (\frac{tan ^{7} ( θ )}{7} + \frac{tan ^{9} ( θ )}{9})

Add a constant to the solution

= 63 (\frac{tan ^{7} ( θ )}{7} + \frac{tan ^{9} ( θ )}{9}) + C

\frac{d}{d x} (x y + ln (\frac{x}{y}))

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = \frac{25}{3}

\int 2 y - 5 d y

x \to + 3 - lim (f (x))

What is the integral of 63sec^4(θ)tan^6(θ) ?
The integral of 63sec^4(θ)tan^6(θ) is 63((tan^7(θ))/7+(tan^9(θ))/9)+C