What is the integral of (sin(x))/(cos^9(x)) ?

The integral of (sin(x))/(cos^9(x)) is 1/8 sec^8(x)+C

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integral of (sin(x))/(cos^9(x))

\int \frac{sin ( x )}{cos ^{9} ( x )} d x

\frac{1}{8} sec^{8} (x) + C

Solution steps

\int \frac{sin ( x )}{cos ^{9} ( x )} d x

Apply u-substitution

: \int - \frac{1}{u ^{9}} d u

= \int - \frac{1}{u ^{9}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{u ^{9}} d u

Apply the Power Rule

: - \frac{1}{8 u ^{8}}

= - (- \frac{1}{8 u ^{8}})

Substitute back

u = cos (x)

= - (- \frac{1}{8 cos ^{8} ( x )})

Simplify

- (- \frac{1}{8 cos ^{8} ( x )}) : \frac{1}{8} sec^{8} (x)

= \frac{1}{8} sec^{8} (x)

Add a constant to the solution

= \frac{1}{8} sec^{8} (x) + C

\int_{- 1}^{7} - x^{2} + 6 x + 7 d x

x \to - 1 lim (t^{2} - x^{2})

\int \frac{( x ^{2} - 3 x - 7 )}{( 2 x + 3 ) ( x + 1 ) ^{2}} d x

\frac{\partial}{\partial x} (x^{3} e^{- 3 y})

t^{3} d x + (x t^{2} + x^{3}) d t = 0

What is the integral of (sin(x))/(cos^9(x)) ?
The integral of (sin(x))/(cos^9(x)) is 1/8 sec^8(x)+C