What is the integral of tan^9(t)sec^4(t) ?

The integral of tan^9(t)sec^4(t) is (tan^{10}(t))/(10)+(tan^{12}(t))/(12)+C

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integral of tan^9(t)sec^4(t)

\int tan^{9} (t) sec^{4} (t) d t

\frac{tan ^{10} ( t )}{10} + \frac{tan ^{12} ( t )}{12} + C

Solution steps

\int tan^{9} (t) sec^{4} (t) d t

Rewrite using trig identities

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

Apply u-substitution

: \int u^{9} (1 + u^{2}) d u

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

Expand

u^{9} (1 + u^{2}) : u^{9} + u^{11}

= \int u^{9} + u^{11} 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

= \int u^{9} d u + \int u^{11} d u

\int u^{9} d u = \frac{u ^{10}}{10}

\int u^{11} d u = \frac{u ^{12}}{12}

= \frac{u ^{10}}{10} + \frac{u ^{12}}{12}

Substitute back

u = tan (t)

= \frac{tan ^{10} ( t )}{10} + \frac{tan ^{12} ( t )}{12}

Add a constant to the solution

= \frac{tan ^{10} ( t )}{10} + \frac{tan ^{12} ( t )}{12} + C

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What is the integral of tan^9(t)sec^4(t) ?
The integral of tan^9(t)sec^4(t) is (tan^{10}(t))/(10)+(tan^{12}(t))/(12)+C