What is the integral of sec^2(θ)tan^9(θ) ?

The integral of sec^2(θ)tan^9(θ) is (tan^{10}(θ))/(10)+C

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integral of sec^2(θ)tan^9(θ)

\int sec^{2} (θ) tan^{9} (θ) d θ

\frac{tan ^{10} ( θ )}{10} + C

Solution steps

\int sec^{2} (θ) tan^{9} (θ) d θ

Rewrite using trig identities

= \int (1 + tan^{2} (θ)) tan^{9} (θ) d θ

Apply u-substitution

: \int u^{9} d u

= \int u^{9} d u

Apply the Power Rule

: \frac{u ^{10}}{10}

= \frac{u ^{10}}{10}

Substitute back

u = tan (θ)

= \frac{tan ^{10} ( θ )}{10}

Add a constant to the solution

= \frac{tan ^{10} ( θ )}{10} + C

\frac{\partial}{\partial x} (2 sec^{2} (2 x - y))

2 x + y^{2} + 2 x y y^{^{'}} = 0

\frac{d}{d x} (x - ln (x + 1))

\int cos^{7} (9 - x) sin (9 - x) d x

v^{'} - 2 = 2 v^{2}

What is the integral of sec^2(θ)tan^9(θ) ?
The integral of sec^2(θ)tan^9(θ) is (tan^{10}(θ))/(10)+C