What is the integral of tan^7(x)sec^6(x) ?

The integral of tan^7(x)sec^6(x) is (tan^8(x))/8+(tan^{10}(x))/5+(tan^{12}(x))/(12)+C

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integral of tan^7(x)sec^6(x)

\int tan^{7} (x) sec^{6} (x) d x

\frac{tan ^{8} ( x )}{8} + \frac{tan ^{10} ( x )}{5} + \frac{tan ^{12} ( x )}{12} + C

Solution steps

\int tan^{7} (x) sec^{6} (x) d x

Rewrite using trig identities

= \int sec^{2} (x) (1 + tan^{2} (x))^{2} tan^{7} (x) d x

Apply u-substitution

: \int u^{7} (1 + u^{2})^{2} d u

= \int u^{7} (1 + u^{2})^{2} d u

Expand

u^{7} (1 + u^{2})^{2} : u^{7} + 2 u^{9} + u^{11}

= \int u^{7} + 2 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^{7} d u + \int 2 u^{9} d u + \int u^{11} d u

\int u^{7} d u = \frac{u ^{8}}{8}

\int 2 u^{9} d u = \frac{u ^{10}}{5}

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

= \frac{u ^{8}}{8} + \frac{u ^{10}}{5} + \frac{u ^{12}}{12}

Substitute back

u = tan (x)

= \frac{tan ^{8} ( x )}{8} + \frac{tan ^{10} ( x )}{5} + \frac{tan ^{12} ( x )}{12}

Add a constant to the solution

= \frac{tan ^{8} ( x )}{8} + \frac{tan ^{10} ( x )}{5} + \frac{tan ^{12} ( x )}{12} + C

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What is the integral of tan^7(x)sec^6(x) ?
The integral of tan^7(x)sec^6(x) is (tan^8(x))/8+(tan^{10}(x))/5+(tan^{12}(x))/(12)+C