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integral of 1/(2sinx + sin2x)
- Thread starter jhp88
- Start date
There are some rather wacky substitutions you can use. I beleiev they're called the Weierstrauss substitutions.
Rewrite your integral using identities:
\(\displaystyle \L\\\frac{1}{2}\int\frac{1}{sin(x)+sin(x)cos(x)}dx\)
Use the substitutions:
\(\displaystyle \L\\sin(x)=\frac{2u}{1+u^{2}}\\cos(x)=\frac{1-u^{2}}{1+u^{2}}\\dx=\frac{2}{1+u^{2}}\\u=tan(\frac{x}{2})\)
\(\displaystyle \L\\\frac{1}{2}\int\frac{1}{\frac{2u}{1+u^{2}}+\frac{2u}{1+u^{2}}\cdot\frac{1-u^{2}}{1+u^{2}}}\cdot\frac{2}{1+u^{2}}du\)
This monster whittles down to:
\(\displaystyle \L\\\frac{1}{2}\int\frac{u^{2}+1}{2u}du\)
Integrating gives:
\(\displaystyle \L\\\frac{2ln(u)+u^{2}}{8}+C\)
Don't forget to sub in \(\displaystyle u=tan(\frac{x}{2})\)
\(\displaystyle \H\\\frac{2ln(tan(\frac{x}{2}))+tan^{2}(\frac{x}{2})}{8}+C\)
Rewrite your integral using identities:
\(\displaystyle \L\\\frac{1}{2}\int\frac{1}{sin(x)+sin(x)cos(x)}dx\)
Use the substitutions:
\(\displaystyle \L\\sin(x)=\frac{2u}{1+u^{2}}\\cos(x)=\frac{1-u^{2}}{1+u^{2}}\\dx=\frac{2}{1+u^{2}}\\u=tan(\frac{x}{2})\)
\(\displaystyle \L\\\frac{1}{2}\int\frac{1}{\frac{2u}{1+u^{2}}+\frac{2u}{1+u^{2}}\cdot\frac{1-u^{2}}{1+u^{2}}}\cdot\frac{2}{1+u^{2}}du\)
This monster whittles down to:
\(\displaystyle \L\\\frac{1}{2}\int\frac{u^{2}+1}{2u}du\)
Integrating gives:
\(\displaystyle \L\\\frac{2ln(u)+u^{2}}{8}+C\)
Don't forget to sub in \(\displaystyle u=tan(\frac{x}{2})\)
\(\displaystyle \H\\\frac{2ln(tan(\frac{x}{2}))+tan^{2}(\frac{x}{2})}{8}+C\)
Hello, jhp88!
Another approach . . .
Multiply top and bottom by \(\displaystyle (1\,-\,\cos x):\)
\(\displaystyle \L\frac{1}{\sin x(1\,+\,\cos x)}\cdot\frac{1\,-\,\cos x}{1\,-\,\cos x} \:=\:\frac{1\,-\,\cos x}{\sin x(1\,-\,\cos^2x)} \:=\:\frac{1\,-\,\cos x}{\sin x\cdot\sin^2x} \:=\:\frac{1\,-\,\cos x}{\sin^3x}\)
. . \(\displaystyle \L=\;\frac{1}{\sin^3x}\,-\,\frac{\cos x}{\sin^3x} \:=\:\csc^3x\,-\,\csc^2x\cot x\)
The integral becomes: \(\displaystyle \:\frac{1}{2}\bigg[\L\int\)\(\displaystyle \csc^3x\,dx \,+\,\L\int\)\(\displaystyle \cot x(-\csc^2x\,dx)\bigg]\)
The first integral is done by parts:\(\displaystyle \L\:\int\)\(\displaystyle \csc^3x\,dx\;=\;\frac{1}{2}\bigg(-\csc x\cot x\,+\,\ln|\csc x\,-\,\cot x|\bigg)\)
In the second integral: let \(\displaystyle u\,=\,\cot x\;\;\Rightarrow\;\;du \,=\,-\csc^2x\,dx\)
. . Substitute: \(\displaystyle \L\:\int\)\(\displaystyle u\,du \;=\;\frac{1}{u^2}\;=\;\frac{1}{2}\cot^2x\)
And we have: \(\displaystyle \:\frac{1}{4}\bigg(\cot^2x\,-\,\csc x\cot x\,+\,\ln|\csc x\,-\,\cot x|\bigg)\,+\,C\)
Another approach . . .
\(\displaystyle \L\int\frac{dx}{2\sin x\,+\,\sin2x}\)
So far I have: \(\displaystyle \L\:\int \frac{dx}{2\sin x\,+\,2\sin x\cos x} \;=\;\frac{1}{2}\int\frac{dx}{\sin x(1\,+\,\cos x)}\;\) . . . a good start!
Multiply top and bottom by \(\displaystyle (1\,-\,\cos x):\)
\(\displaystyle \L\frac{1}{\sin x(1\,+\,\cos x)}\cdot\frac{1\,-\,\cos x}{1\,-\,\cos x} \:=\:\frac{1\,-\,\cos x}{\sin x(1\,-\,\cos^2x)} \:=\:\frac{1\,-\,\cos x}{\sin x\cdot\sin^2x} \:=\:\frac{1\,-\,\cos x}{\sin^3x}\)
. . \(\displaystyle \L=\;\frac{1}{\sin^3x}\,-\,\frac{\cos x}{\sin^3x} \:=\:\csc^3x\,-\,\csc^2x\cot x\)
The integral becomes: \(\displaystyle \:\frac{1}{2}\bigg[\L\int\)\(\displaystyle \csc^3x\,dx \,+\,\L\int\)\(\displaystyle \cot x(-\csc^2x\,dx)\bigg]\)
The first integral is done by parts:\(\displaystyle \L\:\int\)\(\displaystyle \csc^3x\,dx\;=\;\frac{1}{2}\bigg(-\csc x\cot x\,+\,\ln|\csc x\,-\,\cot x|\bigg)\)
In the second integral: let \(\displaystyle u\,=\,\cot x\;\;\Rightarrow\;\;du \,=\,-\csc^2x\,dx\)
. . Substitute: \(\displaystyle \L\:\int\)\(\displaystyle u\,du \;=\;\frac{1}{u^2}\;=\;\frac{1}{2}\cot^2x\)
And we have: \(\displaystyle \:\frac{1}{4}\bigg(\cot^2x\,-\,\csc x\cot x\,+\,\ln|\csc x\,-\,\cot x|\bigg)\,+\,C\)