What is the integral of 5/((sqrt(36-x^2))^3) ?

The integral of 5/((sqrt(36-x^2))^3) is (5x)/(36sqrt(36-x^2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 5/((sqrt(36-x^2))^3)

\int \frac{5}{( 36 - x ^{2} ) ^{3}} d x

\frac{5 x}{36 36 - x ^{2}} + C

Solution steps

\int \frac{5}{( 36 - x ^{2} ) ^{3}} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{( 36 - x ^{2} ) ^{3}} d x

(36 - x^{2})^{3} = (36 - x^{2})^{\frac{3}{2}}

= 5 \cdot \int \frac{1}{( 36 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{36 cos ^{2} ( u )} d u

= 5 \cdot \int \frac{1}{36 cos ^{2} ( u )} d u

Take the constant out:

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

= 5 \cdot \frac{1}{36} \cdot \int \frac{1}{cos ^{2} ( u )} d u

Use the common integral:

\int \frac{1}{cos ^{2} ( u )} d u = tan (u)

= 5 \cdot \frac{1}{36} tan (u)

Substitute back

u = arcsin (\frac{1}{6} x)

= 5 \cdot \frac{1}{36} tan (arcsin (\frac{1}{6} x))

Simplify

5 \cdot \frac{1}{36} tan (arcsin (\frac{1}{6} x)) : \frac{5 x}{36 36 - x ^{2}}

= \frac{5 x}{36 36 - x ^{2}}

Add a constant to the solution

= \frac{5 x}{36 36 - x ^{2}} + C

\int \frac{z ^{4}}{( z ^{2} - 1 ) ^{2}} d z

\int 1000 d x

\int e^{c o s^{2} (x)} sin (2 x) d x

\int \frac{2 x - 3}{x ^{2} - 3 x + 6} d x

\int x^{2} \frac{d}{1} d x

What is the integral of 5/((sqrt(36-x^2))^3) ?
The integral of 5/((sqrt(36-x^2))^3) is (5x)/(36sqrt(36-x^2))+C