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

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

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

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

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

Solution steps

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

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