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

Integrate the given trigonometric function ∫1sin⁡x−sin⁡2xdx\int{\dfrac{1}{\sin x-\sin 2x}dx}.

Explanation

Solution

Hint: Use the formula sin⁡2x=2sin⁡x.cos⁡x\sin 2x=2\sin x.\cos x and simplify the integral. And now use the substitution method. Substitute cos⁡x\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 ∫1sin⁡x−sin⁡2xdx\int{\dfrac{1}{\sin x-\sin 2x}dx}
We know, sin⁡2x=2sin⁡x.cos⁡x\sin 2x=2\sin x.\cos x.
Replacing sin⁡2x\sin 2x with 2sin⁡x.cos⁡x2\sin x.\cos x, we will get,
∫1sin⁡x−sin⁡2xdx=∫1sin⁡x−2sin⁡xcos⁡x\int{\dfrac{1}{\sin x-\sin 2x}dx}=\int{\dfrac{1}{\sin x-2\sin x\cos x}}
Taking sin⁡x\sin x common in denominator, we will get,
∫1sin⁡x−sin⁡2xdx=∫1sin⁡x(1−2cos⁡x)dx\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\sin x in the integral, we will get,
∫1sin⁡x−sin⁡2xdx=∫sin⁡xdxsin⁡2x(1−2cos⁡x)\int{\dfrac{1}{\sin x-\sin 2x}dx}=\int{\dfrac{\sin xdx}{{{\sin }^{2}}x\left( 1-2\cos x \right)}}
Let us assume ∫1sin⁡x−sin⁡2xdx=I\int{\dfrac{1}{\sin x-\sin 2x}dx}=I
⇒I=∫sin⁡xdxsin⁡2x(1−2cos⁡x)..........(1)\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\cos x,
i.e. t=cos⁡xt=\cos x
Differentiating both sides with respect to xx, we will get,
⇒dtdx=dcos⁡xdx\Rightarrow \dfrac{dt}{dx}=\dfrac{d\cos x}{dx}
We know, dcos⁡xdx=−sin⁡x\dfrac{d\cos x}{dx}=-\sin x
⇒dtdx=−sin⁡x\Rightarrow \dfrac{dt}{dx}=-\sin x
Multiplying both sides by dxdx, we will get,
⇒dt=−sin⁡xdx..........(2)\Rightarrow dt=-\sin xdx..........\left( 2 \right)
From equation (1),
⇒I=∫sin⁡xdxsin⁡2x(1−2cos⁡x)\Rightarrow I=\int{\dfrac{\sin xdx}{{{\sin }^{2}}x\left( 1-2\cos x \right)}}
We know,
sin⁡2x+cos⁡2x=1 sin⁡2x=1−cos⁡2x \begin{aligned} & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\\ & {{\sin }^{2}}x=1-{{\cos }^{2}}x \\\ \end{aligned}
Putting sin⁡2x=1−cos⁡2x{{\sin }^{2}}x=1-{{\cos }^{2}}x, we will get,
⇒I=∫sin⁡xdx(1−cos⁡2x)(1−2cos⁡x)\Rightarrow I=\int{\dfrac{\sin xdx}{\left( 1-{{\cos }^{2}}x \right)\left( 1-2\cos x \right)}}
From equation (2),
⇒dt=−sin⁡xdx ⇒sin⁡xdx=−dt \begin{aligned} & \Rightarrow dt=-\sin xdx \\\ & \Rightarrow \sin xdx=-dt \\\ \end{aligned}
Putting sin⁡xdx=−dt\sin xdx=-dt and cos⁡x=t\cos x=t, we will get,
I=∫−dt(1−t2)(1−2t)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=∫dt(1−t2)(2t−1)I=\int{\dfrac{dt}{\left( 1-{{t}^{2}} \right)\left( 2t-1 \right)}}
We know a2−b2=(a−b)(a+b){{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)
Using this, (1−t2)=(1−t)(1+t)\left( 1-{{t}^{2}} \right)=\left( 1-t \right)\left( 1+t \right),
I=∫dt(1−t)(1+t)(2t−1)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 1(1−t)(1+t)(2t−1)=A1+t+B1−t+C2t−1\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,
1(1+t)(1−t)(2t−1)=A(1−t)(2t−1)+B(1+t)(2t−1)+C(1+t)(1−t)(1+t)(1−t)(2t−1)\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 (1−t)(1+t)(2t−1)\left( 1-t \right)\left( 1+t \right)\left( 2t-1 \right), we will get,
1=A(1−t)(2t−1)+B(1+t)(2t−1)+C(1+t)(1−t).......(3)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=1t=1 in equation (3),

& 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{aligned}$$ Putting $t=-1$ in equation (3) $$\begin{aligned} & 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{aligned}$$ Putting $t=\dfrac{1}{2}$ in equation (3), $$\begin{aligned} & 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{aligned}$$ Putting the above value of $$\dfrac{1}{\left( 1-t \right)\left( 1+t \right)\left( 2t-1 \right)}$$ in I, we will get, $I=\int{\left[ \dfrac{A}{1+t}+\dfrac{B}{1-t}+\dfrac{C}{2t-1} \right]}dt$ We know, $\begin{aligned} & \int{\left( f\left( x \right)+g\left( x \right)+h\left( x \right) \right)dx}=\int{f\left( x \right)dx}+\int{g\left( x \right)dx}+\int{h\left( x \right)dx} \\\ & \Rightarrow I=\int{\dfrac{A}{1+t}dt}+\int{\dfrac{B}{1-t}dt}+\int{\dfrac{C}{2t-1}dt} \\\ \end{aligned}$ Taking constants outside the integration, we will get, $I=A\int{\dfrac{1}{1+t}dt}+B\int{\dfrac{1}{1-t}dt}+C\int{\dfrac{1}{2t-1}dt}$ We know, $\int{\dfrac{1}{ax+b}=\dfrac{\ln \left| \left( ax+b \right) \right|}{a}}$ $\Rightarrow I=A\left[ \dfrac{\ln \left| \left( 1+t \right) \right|}{1} \right]+B\left[ \dfrac{\ln \left| \left( 1-t \right) \right|}{-1} \right]+C\left[ \dfrac{\ln \left| \left( 2t-1 \right) \right|}{1} \right]+K$ Putting values of A, B and C, we will get, $$\begin{aligned} & \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{aligned}$$ Now, putting $t=\cos x$, we will get, $$\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 $\ln \left| ax+b \right|+c\ \left( i.e.\ {{\log }_{e}}\ \text{of modulus of }\left( ax+b \right)+c \right)$.