Question

$\int\frac{1}{\sin(x-a)\sin x}dx=$

• A. $\sin a(\log(\sin(x-a)\cdot \csc x))+c$, where c is a constant of integration.
• B. $\csc a(\log(\sin(x-a)\cdot \csc x))+c$, where c is a constant of integration.
• C. $-\sin a(\log(\sin(x-a)\cdot \sin x))+c$, where c is a constant of integration.
• D. $-\csc a(\log(\sin(x-a)\cdot \sin x))+c$, where c is a constant of integration.

Hint:

Logic Tip: The trick depends on the denominator functions. If the denominator has functions of the SAME type (like $\sin \sin$ or $\cos \cos$), multiply and divide by the SINE of the angle difference. If the functions are of DIFFERENT types (like $\sin \cos$), multiply and divide by the COSINE of the angle difference.

Answer 1

The Correct Option is B

Concept:
When the integrand is of the form $\frac{1}{\sin(x-a)\sin(x-b)}$ or $\frac{1}{\cos(x-a)\cos(x-b)}$, multiply and divide the integral by $\sin(A-B)$, where $A$ and $B$ are the angles in the denominator. This allows you to split the numerator using the sine subtraction identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$.
Step 1: Manipulate the integrand by introducing a constant.
Let $I = \int\frac{1}{\sin(x-a)\sin x}dx$. The angles in the denominator are $x$ and $(x-a)$. Their difference is $x - (x-a) = a$. Multiply the numerator and denominator by $\sin a$: $$I = \frac{1}{\sin a} \int \frac{\sin a}{\sin(x-a)\sin x} dx$$
Step 2: Rewrite the numerator and apply the trigonometric identity.
Rewrite the $a$ in the numerator as $(x - (x-a))$: $$I = \csc a \int \frac{\sin(x - (x-a))}{\sin(x-a)\sin x} dx$$ Expand the numerator using $\sin(A-B) = \sin A \cos B - \cos A \sin B$: $$I = \csc a \int \frac{\sin x \cos(x-a) - \cos x \sin(x-a)}{\sin(x-a)\sin x} dx$$
Step 3: Separate the fractions and simplify.
Split the integral into two parts: $$I = \csc a \int \left[ \frac{\sin x \cos(x-a)}{\sin(x-a)\sin x} - \frac{\cos x \sin(x-a)}{\sin(x-a)\sin x} \right] dx$$ Cancel out the common terms: $$I = \csc a \int \left[ \frac{\cos(x-a)}{\sin(x-a)} - \frac{\cos x}{\sin x} \right] dx$$ $$I = \csc a \int \left[ \cot(x-a) - \cot x \right] dx$$
Step 4: Integrate and apply logarithm properties.
Using the standard integral $\int \cot \theta \, d\theta = \log|\sin \theta|$: $$I = \csc a \left[ \log|\sin(x-a)| - \log|\sin x| \right] + c$$ Use the quotient rule for logarithms $\log m - \log n = \log(\frac{m}{n})$: $$I = \csc a \log\left| \frac{\sin(x-a)}{\sin x} \right| + c$$ Since $\frac{1}{\sin x} = \csc x$, rewrite the expression to match the options: $$I = \csc a \log|\sin(x-a) \cdot \csc x| + c$$