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

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

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integral of sec^2(x)tan^9(x)

\int sec^{2} (x) tan^{9} (x) d x

\frac{tan ^{10} ( x )}{10} + C

Solution steps

\int sec^{2} (x) tan^{9} (x) d x

Rewrite using trig identities

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

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 (x)

= \frac{tan ^{10} ( x )}{10}

Add a constant to the solution

= \frac{tan ^{10} ( x )}{10} + C

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{3}}{x ^{2} - 1})

n = 1 \sum \infty \frac{1}{3 ^{n} - 5}

2 + \frac{3}{1 + ( x + 1 ) ^{2}}

\frac{d}{d a} (cos (a) sin (a))

\frac{d}{d x} (- 5 x y (x + y) + 6)

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