This page covers the derivation and use of the sine reduction formula for integration.
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Sine Reduction Formula (where n is a positive integer) |
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\(\displaystyle{\int{\sin^n x~dx} = -\frac{\sin^{n-1}x\cos x}{n}+\frac{n-1}{n}\int{\sin^{n-2}x~dx}}\) |
When you have an integral with only sine where the power is greater than one, you can use the sine reduction formula, repeatedly if necessary, to reduce the power until you end up with either \(\sin x\) or \(\sin^2 x\). Let's derive the formula and then work some practice problems.
Deriving The Sine Reduction Formula
First, we separate out one \(\sin(x)\) term. |
\( \int{\sin^nx~dx} = \int{\sin^{n-1}x\sin x~dx} \) |
Now we use integration by parts |
\(u=\sin^{n-1}x ~~~ \to \) \( ~~~ du=(n-1)\sin^{n-2}x\cos x~dx\) |
\(dv=\sin x~dx ~~~ \to \) \( ~~~ v = -\cos x\) |
\(\int{\sin^nx~dx} = \) \(-\cos x\sin^{n-1}x - \) \(\int{-\cos x(n-1)\sin^{n-2}x\cos x~dx}\) |
Simplify |
\(\int{\sin^nx~dx} = \) \(-\cos x\sin^{n-1}x \) + \((n-1)\int{\sin^{n-2}x\cos^2x~dx}\) |
Now we use the identity \(\cos^2x+\sin^2x=1\) to replace \(\cos^2x\) with \(1-\sin^2x\) in the last integral. |
\(\int{\sin^nx~dx} = \) \(-\cos x\sin^{n-1}x \) + \((n-1)\int{\sin^{n-2}x~(1-\sin^2x)~dx}\) |
Next distribute the \(\sin^{n-2}x\). |
\(\int{\sin^nx~dx} = \) \(-\cos x\sin^{n-1}x \) + \((n-1)\int{\sin^{n-2}x-\sin^nx~dx}\) |
Separate the integral on the right into two integrals. Don't forget that the \((n-1)\) factor needs to be applied to both integrals. |
\(\int{\sin^nx~dx} = \) \(-\cos x\sin^{n-1}x \) + \((n-1)\int{\sin^{n-2}x~dx} - \) \((n-1)\int{\sin^nx~dx}\) |
Now we add \((n-1)\int{\sin^nx~dx} \) to both sides of the equal sign. |
\(\int{\sin^nx~dx} + (n-1)\int{\sin^nx~dx} = \) \(-\cos x\sin^{n-1}x \) + \((n-1)\int{\sin^{n-2}x~dx} \) |
Factor \(\int{\sin^nx~dx}\) on the left and notice that \(1+n-1 = n\). |
\(n\int{\sin^nx~dx} = \) \(-\cos x\sin^{n-1}x \) + \((n-1)\int{\sin^{n-2}x~dx} \) |
Divide both sides by \(n\) to solve for \(\int{\sin^nx~dx}\). |
\(\displaystyle{\int{\sin^nx~dx} = }\) \(\displaystyle{\frac{-\cos x\sin^{n-1}x}{n} }\) + \(\displaystyle{\frac{n-1}{n}\int{\sin^{n-2}x~dx} }\) |
This last equation is the sine reduction formula. Here are a couple of videos showing this derivation in similar ways.
video by Dr Chris Tisdell |
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video by Michael Penn |
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Now let's work some practice problems.
Practice
Unless otherwise instructed, evaluate these integrals directly, then check your answer using the reduction formula.
\(\int{ \sin^2 x ~dx }\)
Problem Statement |
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Evaluate \(\int{ \sin^2 x ~dx }\) directly using trig identities, then check your answer using the reduction formula.
Solution |
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video by The Organic Chemistry Tutor |
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Log in to rate this practice problem and to see it's current rating. |
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\(\int{ \sin^3 x ~dx }\)
Problem Statement |
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Integrate \(\int{ \sin^3 x ~dx }\) directly, then check your answer using the reduction formula.
Solution |
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video by The Organic Chemistry Tutor |
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Log in to rate this practice problem and to see it's current rating. |
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\(\int{ \sin^4 x ~dx }\)
Problem Statement |
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Integrate \(\int{ \sin^4 x ~dx }\) directly, then check your answer using the reduction formula.
Solution |
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video by The Organic Chemistry Tutor |
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Log in to rate this practice problem and to see it's current rating. |
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\(\int{ \sin^5 x ~dx }\)
Problem Statement |
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Integrate \(\int{ \sin^5 x ~dx }\) using u-substitution and trig identities, then check your answer using the reduction formula.
Solution |
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Here are three videos, by three different instructors, solving this problem.
video by Michael Penn |
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video by The Organic Chemistry Tutor |
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video by PatrickJMT |
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Log in to rate this practice problem and to see it's current rating. |
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\(\displaystyle{\int_{0}^{\pi/2}{\sin^7x~dx}}\)
Problem Statement |
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Evaluate \(\displaystyle{\int_{0}^{\pi/2}{\sin^7x~dx}}\) directly, then check your answer using the reduction formula.
Solution |
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video by Dr Chris Tisdell |
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Log in to rate this practice problem and to see it's current rating. |
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Really UNDERSTAND Calculus
basic trig identities |
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\(\sin^2\theta+\cos^2\theta=1\) | \(1+\tan^2\theta=\sec^2\theta\) |
\(\displaystyle{\tan\theta=\frac{\sin\theta}{\cos\theta}}\) | \(\displaystyle{\cot\theta=\frac{\cos\theta}{\sin\theta}}\) |
\(\displaystyle{\sec\theta=\frac{1}{\cos\theta}}\) | \(\displaystyle{\csc\theta=\frac{1}{\sin\theta}}\) |
power reduction (half-angle) formulae |
\(\displaystyle{\sin^2\theta=\frac{1-\cos(2\theta)}{2}}\) | \(\displaystyle{\cos^2\theta=\frac{1+\cos(2\theta)}{2}}\) |
double angle formulae |
\(\sin(2\theta)=2\sin\theta\cos\theta\) | \(\cos(2\theta)=\cos^2\theta-\sin^2\theta\) |
links |
basic trig derivatives | ||
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\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = }\) \(\sec(t)\tan(t) \) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = }\) \( -\csc(t)\cot(t) \) | |
basic trig integrals | ||
\(\int{\sin(x)~dx} = -\cos(x)+C\) | ||
\(\int{\cos(x)~dx} = \sin(x)+C\) | ||
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) | ||
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | ||
\(\int{\sec(x)~dx} = \ln\abs{\sec(x)+\tan(x)}+C\) | ||
\(\int{\csc(x)~dx} = -\ln\abs{\csc(x)+\cot(x)}+C\) | ||
reduction formulae | ||
Reduction Formulas (n is a positive integer) | ||
\(\displaystyle{\int{\sin^n x~dx} = -\frac{\sin^{n-1}x\cos x}{n} + }\) \(\displaystyle{ \frac{n-1}{n}\int{\sin^{n-2}x~dx} }\) | ||
\(\displaystyle{\int{\cos^n x~dx} = \frac{\cos^{n-1}x\sin x}{n} + }\) \(\displaystyle{ \frac{n-1}{n}\int{\cos^{n-2}x~dx}}\) | ||
Reduction Formulas (n is an integer and \(n>1\)) | ||
\(\displaystyle{\int{\tan^n x~dx}= \frac{\tan^{n-1}x}{n-1} - \int{\tan^{n-2}x~dx}}\) | ||
\(\displaystyle{\int{\sec^n x~dx} = \frac{\sec^{n-2}x\tan x}{n-1} + }\) \(\displaystyle{ \frac{n-2}{n-1}\int{\sec^{n-2}x~dx}}\) | ||
links | ||
related topics on other pages |
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external links you may find helpful |
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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Calculus, Better Explained: A Guide To Developing Lasting Intuition |
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Practice Instructions
Unless otherwise instructed, evaluate these integrals directly, then check your answer using the reduction formula.