What is the integral of (sin^7(x))/(cos^4(x)) ?

The integral of (sin^7(x))/(cos^4(x)) is 1/3 sec^3(x)-3sec(x)-3cos(x)+1/3 cos^3(x)+C

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integral of (sin^7(x))/(cos^4(x))

\int \frac{sin ^{7} ( x )}{cos ^{4} ( x )} d x

\frac{1}{3} sec^{3} (x) - 3 sec (x) - 3 cos (x) + \frac{1}{3} cos^{3} (x) + C

Solution steps

\int \frac{sin ^{7} ( x )}{cos ^{4} ( x )} d x

Rewrite using trig identities

= \int \frac{( 1 - cos ^{2} ( x ) ) ^{3} sin ( x )}{cos ^{4} ( x )} d x

Apply u-substitution

: \int - \frac{( 1 - u ^{2} ) ^{3}}{u ^{4}} d u

= \int - \frac{( 1 - u ^{2} ) ^{3}}{u ^{4}} d u

Expand

- \frac{( 1 - u ^{2} ) ^{3}}{u ^{4}} : - \frac{1}{u ^{4}} + \frac{3}{u ^{2}} - 3 + u^{2}

= \int - \frac{1}{u ^{4}} + \frac{3}{u ^{2}} - 3 + u^{2} 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 \frac{1}{u ^{4}} d u + \int \frac{3}{u ^{2}} d u - \int 3 d u + \int u^{2} d u

\int \frac{1}{u ^{4}} d u = - \frac{1}{3 u ^{3}}

\int \frac{3}{u ^{2}} d u = - \frac{3}{u}

\int 3 d u = 3 u

\int u^{2} d u = \frac{u ^{3}}{3}

= - (- \frac{1}{3 u ^{3}}) - \frac{3}{u} - 3 u + \frac{u ^{3}}{3}

Substitute back

u = cos (x)

= - (- \frac{1}{3 cos ^{3} ( x )}) - \frac{3}{cos ( x )} - 3 cos (x) + \frac{cos ^{3} ( x )}{3}

Simplify

- (- \frac{1}{3 cos ^{3} ( x )}) - \frac{3}{cos ( x )} - 3 cos (x) + \frac{cos ^{3} ( x )}{3} : \frac{1}{3} sec^{3} (x) - 3 sec (x) - 3 cos (x) + \frac{1}{3} cos^{3} (x)

= \frac{1}{3} sec^{3} (x) - 3 sec (x) - 3 cos (x) + \frac{1}{3} cos^{3} (x)

Add a constant to the solution

= \frac{1}{3} sec^{3} (x) - 3 sec (x) - 3 cos (x) + \frac{1}{3} cos^{3} (x) + C

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\frac{2}{5 x + 3}, at x = - 1

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

\int cos^{3} (\frac{x}{9}) d x

What is the integral of (sin^7(x))/(cos^4(x)) ?
The integral of (sin^7(x))/(cos^4(x)) is 1/3 sec^3(x)-3sec(x)-3cos(x)+1/3 cos^3(x)+C