What is the integral of (sin(t)cos^{-3}(t)) ?

The integral of (sin(t)cos^{-3}(t)) is 1/2 sec^2(t)+C

Popular >

integral of (sin(t)cos^{-3}(t))

\int (sin (t) cos^{- 3} (t)) d t

\frac{1}{2} sec^{2} (t) + C

Solution steps

\int sin (t) cos^{- 3} (t) d t

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

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

Substitute back

u = cos (t)

= - (- \frac{1}{2 cos ^{2} ( t )})

Simplify

- (- \frac{1}{2 cos ^{2} ( t )}) : \frac{1}{2} sec^{2} (t)

= \frac{1}{2} sec^{2} (t)

Add a constant to the solution

= \frac{1}{2} sec^{2} (t) + C

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

\int \frac{y ln ( y )}{1 + x ^{2}} d x

\int 2 x (x^{2} + 8)^{5} d x

\int arctan (2 x - 1) d x

\int \frac{10}{( x - 1 ) ^{2}} d x

What is the integral of (sin(t)cos^{-3}(t)) ?
The integral of (sin(t)cos^{-3}(t)) is 1/2 sec^2(t)+C