Question

Integrate the given trigonometric function \[\int{\dfrac{1}{\sin x-\sin 2x}dx}\].

Answer 1

Hint: Use the formula $\sin 2x=2\sin x.\cos x$ and simplify the integral. And now use the substitution method. Substitute $\cos x$ with the variable ‘t’. You will get a polynomial in ‘t’. Then use partial fractions method for integrating.

Complete step-by-step solution -
We have to find \[\int{\dfrac{1}{\sin x-\sin 2x}dx}\]
We know, $\sin 2x=2\sin x.\cos x$.
Replacing $\sin 2x$ with $2\sin x.\cos x$, we will get,
\[\int{\dfrac{1}{\sin x-\sin 2x}dx}=\int{\dfrac{1}{\sin x-2\sin x\cos x}}\]
Taking $\sin x$ common in denominator, we will get,
\[\int{\dfrac{1}{\sin x-\sin 2x}dx}=\int{\dfrac{1}{\sin x\left( 1-2\cos x \right)}}dx\]
On multiplying and dividing by $\sin x$ in the integral, we will get,
\[\int{\dfrac{1}{\sin x-\sin 2x}dx}=\int{\dfrac{\sin xdx}{{{\sin }^{2}}x\left( 1-2\cos x \right)}}\]
Let us assume \[\int{\dfrac{1}{\sin x-\sin 2x}dx}=I\]
\[\Rightarrow I=\int{\dfrac{\sin xdx}{{{\sin }^{2}}x\left( 1-2\cos x \right)}}..........\left( 1 \right)\]
Let us substitute a variable ‘t’ for $\cos x$,
i.e. $t=\cos x$
Differentiating both sides with respect to $x$, we will get,
$\Rightarrow \dfrac{dt}{dx}=\dfrac{d\cos x}{dx}$
We know, $\dfrac{d\cos x}{dx}=-\sin x$
$\Rightarrow \dfrac{dt}{dx}=-\sin x$
Multiplying both sides by $dx$, we will get,
$\Rightarrow dt=-\sin xdx..........\left( 2 \right)$
From equation (1),
\[\Rightarrow I=\int{\dfrac{\sin xdx}{{{\sin }^{2}}x\left( 1-2\cos x \right)}}\]
We know,
$\begin{align}
  & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\
 & {{\sin }^{2}}x=1-{{\cos }^{2}}x \\
\end{align}$
Putting ${{\sin }^{2}}x=1-{{\cos }^{2}}x$, we will get,
\[\Rightarrow I=\int{\dfrac{\sin xdx}{\left( 1-{{\cos }^{2}}x \right)\left( 1-2\cos x \right)}}\]
From equation (2),
$\begin{align}
  & \Rightarrow dt=-\sin xdx \\
 & \Rightarrow \sin xdx=-dt \\
\end{align}$
Putting $\sin xdx=-dt$ and $\cos x=t$, we will get,
\[I=\int{\dfrac{-dt}{\left( 1-{{t}^{2}} \right)\left( 1-2t \right)}}\]
 Multiplying both numerator and denominator by -1 in integral, we will get,
\[I=\int{\dfrac{dt}{\left( 1-{{t}^{2}} \right)\left( 2t-1 \right)}}\]
We know ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
Using this, $\left( 1-{{t}^{2}} \right)=\left( 1-t \right)\left( 1+t \right)$,
\[I=\int{\dfrac{dt}{\left( 1-t \right)\left( 1+t \right)\left( 2t-1 \right)}}\]
Now, let us use partial fraction for further solving,
Let \[\dfrac{1}{\left( 1-t \right)\left( 1+t \right)\left( 2t-1 \right)}=\dfrac{A}{1+t}+\dfrac{B}{1-t}+\dfrac{C}{2t-1}\]
Taking LCM and adding in RHS, we will get,
\[\dfrac{1}{\left( 1+t \right)\left( 1-t \right)\left( 2t-1 \right)}=\dfrac{A\left( 1-t \right)\left( 2t-1 \right)+B\left( 1+t \right)\left( 2t-1 \right)+C\left( 1+t \right)\left( 1-t \right)}{\left( 1+t \right)\left( 1-t \right)\left( 2t-1 \right)}\]
Dividing both sides of equation by \[\left( 1-t \right)\left( 1+t \right)\left( 2t-1 \right)\], we will get,
\[1=A\left( 1-t \right)\left( 2t-1 \right)+B\left( 1+t \right)\left( 2t-1 \right)+C\left( 1+t \right)\left( 1-t \right).......\left( 3 \right)\]
Putting $t=1$ in equation (3),
\[\begin{align}
  & 1=A\left( 0 \right)+B\left( 2 \right)\left( 1 \right)+C\left( 0 \right) \\
 & \Rightarrow 1=2B \\
 & \Rightarrow B=\dfrac{1}{2} \\
\end{align}\]
Putting $t=-1$ in equation (3)
\[\begin{align}
  & 1=A\left( 2 \right)\left( -3 \right)+B\left( 0 \right)+C\left( 0 \right) \\
 & \Rightarrow 1=-6A \\
 & \Rightarrow A=\dfrac{-1}{6} \\
\end{align}\]
Putting $t=\dfrac{1}{2}$ in equation (3),
\[\begin{align}
  & 1=A\left( 1-\dfrac{1}{2} \right)\left( 0 \right)+B\left( 1+\dfrac{1}{2} \right)\left( 0 \right)+C\left( 1+\dfrac{1}{2} \right)\left( 1-\dfrac{1}{2} \right) \\
 & \Rightarrow 1=C\left( \dfrac{3}{2} \right)\left( \dfrac{1}{2} \right) \\
 & \Rightarrow 1=\dfrac{3C}{4} \\
 & \Rightarrow C=\dfrac{4}{3} \\
 & \Rightarrow \dfrac{1}{\left( 1-t \right)\left( 1+t \right)\left( 2t-1 \right)}=\dfrac{\dfrac{-1}{6}}{1+t}+\dfrac{\dfrac{1}{2}}{1-t}+\dfrac{\dfrac{4}{3}}{2t-1} \\
\end{align}\]
Putting the above value of \[\dfrac{1}{\left( 1-t \right)\left( 1+t \right)\left( 2t-1 \right)}\] in I, we will get,
$I=\int{\left[ \dfrac{A}{1+t}+\dfrac{B}{1-t}+\dfrac{C}{2t-1} \right]}dt$
We know,
$\begin{align}
  & \int{\left( f\left( x \right)+g\left( x \right)+h\left( x \right) \right)dx}=\int{f\left( x \right)dx}+\int{g\left( x \right)dx}+\int{h\left( x \right)dx} \\
 & \Rightarrow I=\int{\dfrac{A}{1+t}dt}+\int{\dfrac{B}{1-t}dt}+\int{\dfrac{C}{2t-1}dt} \\
\end{align}$
Taking constants outside the integration, we will get,
$I=A\int{\dfrac{1}{1+t}dt}+B\int{\dfrac{1}{1-t}dt}+C\int{\dfrac{1}{2t-1}dt}$
We know, $\int{\dfrac{1}{ax+b}=\dfrac{\ln \left| \left( ax+b \right) \right|}{a}}$
$\Rightarrow I=A\left[ \dfrac{\ln \left| \left( 1+t \right) \right|}{1} \right]+B\left[ \dfrac{\ln \left| \left( 1-t \right) \right|}{-1} \right]+C\left[ \dfrac{\ln \left| \left( 2t-1 \right) \right|}{1} \right]+K$
Putting values of A, B and C, we will get,
\[\begin{align}
  & \Rightarrow I=\dfrac{-1}{6}\left( \ln \left| 1+t \right| \right)+\dfrac{1}{2}\left( \dfrac{\ln \left| \left( 1-t \right) \right|}{-1} \right)+\dfrac{4}{3}\left( \dfrac{\ln \left| \left( 2t-1 \right) \right|}{2} \right)+K \\
 & \Rightarrow I=\dfrac{-1}{6}\left( \ln \left| 1+t \right| \right)-\dfrac{1}{2}\left( \ln \left| \left( 1-t \right) \right| \right)+\dfrac{2}{3}\left( \ln \left| \left( 2t-1 \right) \right| \right)+K \\
\end{align}\]
Now, putting $t=\cos x$, we will get,
\[\Rightarrow I=\dfrac{-1}{6}\ln \left| 1+\cos x \right|-\dfrac{1}{2}\ln \left| \left( 1-\cos x \right) \right|+\dfrac{2}{3}\ln \left| \left( 2\cos x-1 \right) \right|+K\]
Where K is a constant of integration.

Note: Don’t forget to add ‘K’ (the constant of integration). In the question of indefinite integral, it is necessary to add a constant of integration in the answer. Don’t forget to use modulus in the integration of $\dfrac{1}{ax+b}$. Integration of $\dfrac{1}{ax+b}$ with $dx$ is equal to $\ln \left| ax+b \right|+c\ \left( i.e.\ {{\log }_{e}}\ \text{of modulus of }\left( ax+b \right)+c \right)$.