Question

Evaluate the following integral $\int{\dfrac{dx}{\sin x\sin \left( x+\alpha \right)}}$.$$$$
A.\operatorname{cosec}\alpha \ln \left| \dfrac{\sin x}{\sin \left( x+\alpha \right)} \right|+c$$$$$ B. \operatorname{cosec}\alpha \ln \left| \dfrac{\sin \left( x+\alpha \right)}{\sin x} \right|+c C. $\operatorname{cosec}\alpha \ln \sec \left| \dfrac{\sec x}{\sec \left( x+\alpha \right)} \right|+c
D. $\operatorname{cosec}\alpha \ln \sec \left| \dfrac{\sec \left( x+\alpha \right)}{\sec x} \right|+c$$$$$

Answer 1

We multiply and divide $\sin \alpha$ with integrand. We add and subtract $x$ with $\sin \alpha$ in the numerator. We use a difference of angle formula that for two angles $A,B\left( A>B \right)$ that is $\sin \left( A+B \right)=\sin A+\cos B-\cos A\sin B$ and convert the integrand into cot function. We use know the standard indefinite integral of cotangent function $\int{\cot xdx}=\ln \left| \sin x \right|+c$ and then the logarithmic identity of quotient $\ln \left( \dfrac{a}{b} \right)=\ln a-\ln b$ to find the answer. $$$$

Complete step-by-step answer:
We are given in the question the following integral to evaluate$\int{\dfrac{dx}{\sin x\sin \left( x+\alpha \right)}}$. We see the answers in the options and find that in all the options $\operatorname{cosec}\alpha$ is multiplied. We know the reciprocal relation of trigonometric ratios that $\operatorname{cosec}\alpha =\dfrac{1}{\sin \alpha }$.
We keep it in our mind and multiply and divide $\sin \alpha$ in the numerator and denominator of the integrand to have;

& \int{\dfrac{dx}{\sin x\sin \left( x+\alpha \right)}}\times \dfrac{\sin \alpha }{\sin \alpha } \\\ & \Rightarrow \dfrac{1}{\sin \alpha }\int{\dfrac{\sin \alpha dx}{\sin x\sin \left( x+\alpha \right)}} \\\ \end{aligned}$$ We add and subtract $x$ with $\alpha $in the numerator of the integrand in the above step to have; $$\Rightarrow \dfrac{1}{\sin \alpha }\int{\dfrac{\sin \left( x+\alpha -x \right)dx}{\sin x\sin \left( x+\alpha \right)}}$$ We use sine difference of angle formula for $A=x+\alpha ,B=x$ in the numerator of the integrand in the above step to have; $$\Rightarrow \dfrac{1}{\sin \alpha }\int{\dfrac{\sin \left( x+\alpha \right)\cos x-\cos \left( x+\alpha \right)\sin xdx}{\sin x\sin \left( x+\alpha \right)}}$$ We separate the integrand into two terms and use sum rule of integration to have; $$\Rightarrow \dfrac{1}{\sin \alpha }\left( \int{\dfrac{\sin \left( x+\alpha \right)\cos x}{\sin x\sin \left( x+\alpha \right)}}dx-\int{\dfrac{\cos \left( x+\alpha \right)\sin x}{\sin x\sin \left( x+\alpha \right)}dx} \right)$$ We divide $\sin \left( x+\alpha \right)$ with the numerator and denominator of the first integrand and divide $\sin x$ with the numerator and denominator of the first integrand in the above step to have; $$\Rightarrow \dfrac{1}{\sin \alpha }\left( \int{\dfrac{\cos x}{\sin x}}dx-\int{\dfrac{\cos \left( x+\alpha \right)}{\sin \left( x+\alpha \right)}}dx \right)$$ We convert sine and cosine into cotangent using the trigonometric identity $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ for $\theta =x$ in the first integrand and $\theta =x+\alpha $ in the second integrand to have; $$\Rightarrow \dfrac{1}{\sin \alpha }\left( \int{\cot x}dx-\int{\cot \left( x+\alpha \right)}dx \right)$$ We use the standard integral of cotangent function for both the integrands in the above step to have; $$\Rightarrow \dfrac{1}{\sin \alpha }\left( \ln \left| \sin x \right|-\ln \left| \sin \left( x+\alpha \right) \right| \right)+c$$ Here $c$ is an arbitrary integration constant . We use the logarithmic identity of quotient $\ln \left( \dfrac{a}{b} \right)=\ln a-\ln b$ in the above step to have; $$\begin{aligned} & \Rightarrow \dfrac{1}{\sin \alpha }\left( \ln \left| \dfrac{\sin x}{\sin \left( x+\alpha \right)} \right| \right)+c \\\ & \Rightarrow \operatorname{cosec}\alpha \left( \ln \left| \dfrac{\sin x}{\sin \left( x+\alpha \right)} \right| \right)+c \\\ \end{aligned}$$ **So, the correct answer is “Option A”.** **Note:** We note that since in the question $\sin x,\sin \left( x+\alpha \right)$ are well defined $x,x+\alpha $ cannot be an integral multiple of $\pi $ then $\alpha $ cannot be integral multiple of $\pi $ and hence $\operatorname{cosec}\alpha $is well defined in our answer. We must be careful of the confusion between standard definite integral of cotangent function from tangent function which is given by $\int{\tan xdx}=-\ln \left( \cos x \right)+c$.