What is the integral of sec^5(y)tan^3(y) ?

The integral of sec^5(y)tan^3(y) is (sec^7(y))/7-(sec^5(y))/5+C

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integral of sec^5(y)tan^3(y)

\int sec^{5} (y) tan^{3} (y) d y

\frac{sec ^{7} ( y )}{7} - \frac{sec ^{5} ( y )}{5} + C

Solution steps

\int sec^{5} (y) tan^{3} (y) d y

Rewrite using trig identities

= \int (- 1 + sec^{2} (y)) tan (y) sec^{5} (y) d y

Apply u-substitution

: \int u^{4} (u^{2} - 1) d u

= \int u^{4} (u^{2} - 1) d u

Expand

u^{4} (u^{2} - 1) : u^{6} - u^{4}

= \int u^{6} - u^{4} 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^{6} d u - \int u^{4} d u

\int u^{6} d u = \frac{u ^{7}}{7}

\int u^{4} d u = \frac{u ^{5}}{5}

= \frac{u ^{7}}{7} - \frac{u ^{5}}{5}

Substitute back

u = sec (y)

= \frac{sec ^{7} ( y )}{7} - \frac{sec ^{5} ( y )}{5}

Add a constant to the solution

= \frac{sec ^{7} ( y )}{7} - \frac{sec ^{5} ( y )}{5} + C

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What is the integral of sec^5(y)tan^3(y) ?
The integral of sec^5(y)tan^3(y) is (sec^7(y))/7-(sec^5(y))/5+C