Integral $\int \frac {1}{\sin^3(x)} dx$

Question

I need to calculate the following integral for my homework, but I dont know how. If someone show me step by step solution I would really appreciate it. $$\int \frac {1}{\sin^3(x)} dx$$

Answer 1

There is a standard first-year calculus response. Rewrite the integrand as $\dfrac{\sin x}{\sin^4 x}=\dfrac{\sin x}{(1-\cos^2 x)^2}$. The substitution $u=\cos x$ leaves us integrating $-\dfrac{1}{(1-u^2)^2}$. Now partial fractions.

There are in many cases more efficient procedures, but one can in principle handle in this way all $$\int \sin^m x\cos^n x\,dx,$$ where $m$ and $n$ are integers and at least one of $m$ and $n$ is odd.

Answer 2

Substitute $u=\tan\frac{x}{2}$. Then $\frac{x}{2}=\arctan u\Rightarrow x=2\arctan u$. This means $$\frac{dx}{du}=\frac{2}{1+u^2}$$ In addition, $$\sin x=\frac{2u}{u^2+1}$$ Then, $$\int\frac{1}{\sin^3x}dx=\frac{2}{8}\int\frac{(u^2+1)^2}{u^3}du=\frac{1}{4}\int u+\frac{2}{u}+\frac{1}{u^3}du$$ which I think you know how to solve.

What I used here is the classic but always useful Tangent half-angle formula

Answer 3

integral csc^3(x) dx

Use the reduction formula,

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-cos(x) sin^2(x)^((m-1)/2) csc^(m-1)(x) 2F1(1/2, (m+1)/2, 3/2, cos^2(x)) = -(cos(x) csc^(m-1)(x))/(m-1) + (m-2)/(m-1)-cos(x) sin^2(x)^((m-3)/2) csc^(m-3)(x) 2F1(1/2, (m-1)/2, 3/2, cos^2(x)), where m = 3:

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= 1/2 integral csc(x) dx-1/2 cot(x) csc(x)

The integral of csc(x) is -log(cot(x)+csc(x)): = -1/2 (cot(x) csc(x))-1/2 log(cot(x)+csc(x))+constant

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Factor the answer a different way:

Answer: = 1/2 (-cot(x) csc(x)-log(cot(x)+csc(x)))+constant

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Sorry for the mess.

Answer 4

Make the change of variables $ \csc(x) = u $, which implies

$$ \int \csc(\theta)^3 d \theta = -\int \frac{{u^2}}{\sqrt{u^2-1}} du \,. $$

Following it with the change of variables $ u = \sin(t) $ yields

$$ -\int \frac{{u^2}}{\sqrt{u^2-1}} du = i \int \sin(t)^2\,dz= \dots. $$

Note: You can use integration by parts to evaluate the integral

$$ -\int \frac{{u^2}}{\sqrt{u^2-1}} du . $$