What is the integral of (sin^3(x))/(cos^7(x)) ?

The integral of (sin^3(x))/(cos^7(x)) is 1/6 sec^6(x)-1/4 sec^4(x)+C

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integral of (sin^3(x))/(cos^7(x))

\int \frac{sin ^{3} ( x )}{cos ^{7} ( x )} d x

\frac{1}{6} sec^{6} (x) - \frac{1}{4} sec^{4} (x) + C

Solution steps

\int \frac{sin ^{3} ( x )}{cos ^{7} ( x )} d x

Rewrite using trig identities

= \int \frac{( 1 - cos ^{2} ( x ) ) sin ( x )}{cos ^{7} ( x )} d x

Apply u-substitution

: \int - \frac{1 - u ^{2}}{u ^{7}} d u

= \int - \frac{1 - u ^{2}}{u ^{7}} d u

Expand

- \frac{1 - u ^{2}}{u ^{7}} : - \frac{1}{u ^{7}} + \frac{1}{u ^{5}}

= \int - \frac{1}{u ^{7}} + \frac{1}{u ^{5}} 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 ^{7}} d u + \int \frac{1}{u ^{5}} d u

\int \frac{1}{u ^{7}} d u = - \frac{1}{6 u ^{6}}

\int \frac{1}{u ^{5}} d u = - \frac{1}{4 u ^{4}}

= - (- \frac{1}{6 u ^{6}}) - \frac{1}{4 u ^{4}}

Substitute back

u = cos (x)

= - (- \frac{1}{6 cos ^{6} ( x )}) - \frac{1}{4 cos ^{4} ( x )}

Simplify

- (- \frac{1}{6 cos ^{6} ( x )}) - \frac{1}{4 cos ^{4} ( x )} : \frac{1}{6} sec^{6} (x) - \frac{1}{4} sec^{4} (x)

= \frac{1}{6} sec^{6} (x) - \frac{1}{4} sec^{4} (x)

Add a constant to the solution

= \frac{1}{6} sec^{6} (x) - \frac{1}{4} sec^{4} (x) + C

\frac{d}{d x} (\frac{4}{3} π x^{3})

\int_{1}^{100} \frac{37}{x ^{3}} d x

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

f (x) = 4 x + 3

x - \frac{π}{2}, 1

What is the integral of (sin^3(x))/(cos^7(x)) ?
The integral of (sin^3(x))/(cos^7(x)) is 1/6 sec^6(x)-1/4 sec^4(x)+C