What is the integral of (sin^7(x))/(cos^2(x)) ?

The integral of (sin^7(x))/(cos^2(x)) is sec(x)+3cos(x)-cos^3(x)+1/5 cos^5(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sin^7(x))/(cos^2(x))

integral

\frac{sin ^{7} ( x )}{cos ^{2} ( x )}

sec (x) + 3 cos (x) - cos^{3} (x) + \frac{1}{5} cos^{5} (x) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

\int 3 d u = 3 u

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

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

= - (- \frac{1}{u}) + 3 u - u^{3} + \frac{u ^{5}}{5}

Substitute back

u = cos (x)

= - (- \frac{1}{cos ( x )}) + 3 cos (x) - cos^{3} (x) + \frac{cos ^{5} ( x )}{5}

Simplify

- (- \frac{1}{cos ( x )}) + 3 cos (x) - cos^{3} (x) + \frac{cos ^{5} ( x )}{5} : sec (x) + 3 cos (x) - cos^{3} (x) + \frac{1}{5} cos^{5} (x)

= sec (x) + 3 cos (x) - cos^{3} (x) + \frac{1}{5} cos^{5} (x)

Add a constant to the solution

= sec (x) + 3 cos (x) - cos^{3} (x) + \frac{1}{5} cos^{5} (x) + C

x^{2} - 2 x + 1

\frac{1}{( 1 + r ^{2} ) ^{2}}

\frac{1}{( 1 + ( x ) ^{\frac{1}{2}} )}

\frac{( 1 - x )}{( 1 + x )}

\frac{1}{( 2 + sin ( 3 x ) )}

What is the integral of (sin^7(x))/(cos^2(x)) ?
The integral of (sin^7(x))/(cos^2(x)) is sec(x)+3cos(x)-cos^3(x)+1/5 cos^5(x)+C