What is the integral of (csc^2(t))/(1+9cot^2(t)) ?

The integral of (csc^2(t))/(1+9cot^2(t)) is 1/3 arctan((tan(t))/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (csc^2(t))/(1+9cot^2(t))

\int \frac{csc ^{2} ( t )}{1 + 9 cot ^{2} ( t )} d t

\frac{1}{3} arctan (\frac{tan ( t )}{3}) + C

Solution steps

\int \frac{csc ^{2} ( t )}{1 + 9 cot ^{2} ( t )} d t

Rewrite using trig identities

= \int \frac{1}{1 + 8 cos ^{2} ( t )} d t

Apply Trigonometric Substitution

: \int \frac{1}{u ^{2} + 9} d u

= \int \frac{1}{u ^{2} + 9} d u

Apply Integral Substitution

= \int \frac{1}{3 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{3} arctan (v)

Substitute back

= \frac{1}{3} arctan (\frac{tan ( t )}{3})

Add a constant to the solution

= \frac{1}{3} arctan (\frac{tan ( t )}{3}) + C

\int 25 - 5 x^{2} d x

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

\int \frac{5}{e ^{2 x}} d x

\int \frac{1}{16 - 6 x - x ^{2}} d x

\int x^{- \frac{7}{8}} d x

What is the integral of (csc^2(t))/(1+9cot^2(t)) ?
The integral of (csc^2(t))/(1+9cot^2(t)) is 1/3 arctan((tan(t))/3)+C