What is the integral of 1/(t^2sqrt(441-t^2)) ?

The integral of 1/(t^2sqrt(441-t^2)) is -(sqrt(441-t^2))/(441t)+C

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integral of 1/(t^2sqrt(441-t^2))

\int \frac{1}{t ^{2} 441 - t ^{2}} d t

- \frac{441 - t ^{2}}{441 t} + C

Solution steps

\int \frac{1}{t ^{2} 441 - t ^{2}} d t

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{441 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{441 cos ( u ) sin ( u )} d u

Take the constant out:

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

= \frac{1}{441} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

\frac{cot ( u )}{cos ( u ) sin ( u )} : csc^{2} (u)

= \frac{1}{441} \cdot \int csc^{2} (u) d u

Use the common integral:

\int csc^{2} (u) d u = - cot (u)

= \frac{1}{441} (- cot (u))

Substitute back

u = arcsin (\frac{1}{21} t)

= \frac{1}{441} (- cot (arcsin (\frac{1}{21} t)))

Simplify

\frac{1}{441} (- cot (arcsin (\frac{1}{21} t))) : - \frac{441 - t ^{2}}{441 t}

= - \frac{441 - t ^{2}}{441 t}

Add a constant to the solution

= - \frac{441 - t ^{2}}{441 t} + C

\int (4 x^{5} + 1) x^{4} d x

\frac{\partial}{\partial x} (cos^{3} (x^{7} y^{3}))

\frac{\partial}{\partial x} (9 x^{2} y - 3 x^{5} y)

\frac{\partial}{\partial x} (x \cdot y \cdot z)

x \to 4 + lim (\frac{( 7 x )}{x - 4})

What is the integral of 1/(t^2sqrt(441-t^2)) ?
The integral of 1/(t^2sqrt(441-t^2)) is -(sqrt(441-t^2))/(441t)+C