Answer on Question #50934 – Math – Integral Calculus
Integrate with respect to x x x ∫ csc 2 x d x \int \csc 2x \, dx ∫ csc 2 x d x
Solution.
First we use Trigonometric Identity, namely Reciprocal Identity, namely: csc x = 1 sin x \csc x = \frac{1}{\sin x} csc x = s i n x 1
According to Reciprocal Identity csc 2 x = 1 sin 2 x \csc 2x = \frac{1}{\sin 2x} csc 2 x = s i n 2 x 1
I = ∫ csc 2 x d x = ∫ 1 sin 2 x d x ⇒ I = \int \csc 2 x \, dx = \int \frac{1}{\sin 2x} \, dx \Rightarrow I = ∫ csc 2 x d x = ∫ sin 2 x 1 d x ⇒
**Step 1.** Let's multiply and divide the integrand, i.e. the function that is to be integrated, 1 sin 2 x \frac{1}{\sin 2x} s i n 2 x 1 sin 2 x \sin 2x sin 2 x
I = ∫ csc 2 x d x = ∫ 1 sin 2 x d x = ∫ sin 2 x sin 2 2 x d x = ∣ Pythagorean identity ∣ sin 2 2 x = 1 − cos 2 2 x = ∫ sin 2 x 1 − cos 2 2 x d x = = − 1 2 ∫ d cos 2 x ( 1 − cos 2 2 x ) = ∣ t = cos 2 x ∣ = − 1 2 ∫ d t ( 1 − t 2 ) = − 1 2 ∫ d t ( 1 − t 2 ) ⇒ \begin{array}{l}
I = \int \csc 2 x \, dx = \int \frac{1}{\sin 2x} \, dx = \int \frac{\sin 2x}{\sin^2 2x} \, dx = \left| \text{Pythagorean identity} \right| \sin^2 2x = 1 - \cos^2 2x = \int \frac{\sin 2x}{1 - \cos^2 2x} \, dx = \\
= -\frac{1}{2} \int \frac{d \cos 2x}{(1 - \cos^2 2x)} = |t = \cos 2x| = -\frac{1}{2} \int \frac{dt}{(1 - t^2)} = -\frac{1}{2} \int \frac{dt}{(1 - t^2)} \Rightarrow \\
\end{array} I = ∫ csc 2 x d x = ∫ s i n 2 x 1 d x = ∫ s i n 2 2 x s i n 2 x d x = ∣ Pythagorean identity ∣ sin 2 2 x = 1 − cos 2 2 x = ∫ 1 − c o s 2 2 x s i n 2 x d x = = − 2 1 ∫ ( 1 − c o s 2 2 x ) d c o s 2 x = ∣ t = cos 2 x ∣ = − 2 1 ∫ ( 1 − t 2 ) d t = − 2 1 ∫ ( 1 − t 2 ) d t ⇒
**Step 2.** Let's use the method of Partial Fraction Decomposition.
1) Factor the denominator: ( 1 − t 2 ) = ( 1 − t ) ( 1 + t ) (1 - t^2) = (1 - t)(1 + t) ( 1 − t 2 ) = ( 1 − t ) ( 1 + t )
2) Write one partial fraction for each of those factors:
3) 1 ( 1 − t 2 ) = A ( 1 − t ) + B ( 1 + t ) = A ( 1 + t ) + B ( 1 − t ) ( 1 − t 2 ) \frac{1}{(1 - t^2)} = \frac{A}{(1 - t)} + \frac{B}{(1 + t)} = \frac{A(1 + t) + B(1 - t)}{(1 - t^2)} ( 1 − t 2 ) 1 = ( 1 − t ) A + ( 1 + t ) B = ( 1 − t 2 ) A ( 1 + t ) + B ( 1 − t )
4) Multiply through by the denominator so we no longer have fractions. A ( 1 + t ) + B ( 1 − t ) = 1 A(1 + t) + B(1 - t) = 1 A ( 1 + t ) + B ( 1 − t ) = 1
5) Find unknown coefficients A A A B B B
{ t = 1 A ( 1 + 1 ) + B ( 1 − 1 ) = 1 t = − 1 A ( 1 + t ) + B ( 1 − t ) = 1 ⇒ { 2 A = 1 2 B = 1 ⇒ { A = 1 2 B = 1 2 \left\{
\begin{array}{l}
t = 1 \quad A(1 + 1) + B(1 - 1) = 1 \\
t = -1 \quad A(1 + t) + B(1 - t) = 1
\end{array}
\right.
\Rightarrow
\left\{
\begin{array}{l}
2A = 1 \\
2B = 1
\end{array}
\right.
\Rightarrow
\left\{
\begin{array}{l}
A = \frac{1}{2} \\
B = \frac{1}{2}
\end{array}
\right. { t = 1 A ( 1 + 1 ) + B ( 1 − 1 ) = 1 t = − 1 A ( 1 + t ) + B ( 1 − t ) = 1 ⇒ { 2 A = 1 2 B = 1 ⇒ { A = 2 1 B = 2 1
**Step 3.**
I = − 1 2 ∫ d t ( 1 − t 2 ) = − 1 4 ∫ d t 1 − t − 1 4 ∫ d t 1 + t = + 1 4 ln ∣ 1 − t ∣ − 1 4 ln ∣ 1 + t ∣ + c = 1 4 ln ∣ 1 − t 1 + t ∣ + c ⇒ I = -\frac{1}{2} \int \frac{dt}{(1 - t^2)} = -\frac{1}{4} \int \frac{dt}{1 - t} - \frac{1}{4} \int \frac{dt}{1 + t} = +\frac{1}{4} \ln |1 - t| - \frac{1}{4} \ln |1 + t| + c = \frac{1}{4} \ln \left| \frac{1 - t}{1 + t} \right| + c \Rightarrow I = − 2 1 ∫ ( 1 − t 2 ) d t = − 4 1 ∫ 1 − t d t − 4 1 ∫ 1 + t d t = + 4 1 ln ∣1 − t ∣ − 4 1 ln ∣1 + t ∣ + c = 4 1 ln ∣ ∣ 1 + t 1 − t ∣ ∣ + c ⇒
**Step 4.** Now let's apply inverse substitution of variables I = 1 4 ln ∣ 1 − cos 2 x 1 + cos 2 x ∣ + c I = \frac{1}{4} \ln \left| \frac{1 - \cos 2x}{1 + \cos 2x} \right| + c I = 4 1 ln ∣ ∣ 1 + c o s 2 x 1 − c o s 2 x ∣ ∣ + c c c c
**Answer:** I = 1 4 ln ∣ 1 − cos 2 x 1 + cos 2 x ∣ + c I = \frac{1}{4} \ln \left| \frac{1 - \cos 2x}{1 + \cos 2x} \right| + c I = 4 1 ln ∣ ∣ 1 + c o s 2 x 1 − c o s 2 x ∣ ∣ + c
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#339914 on Dec 2023