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# $\Large \int\frac{dx}{\sin(x-a)\sin(x-b)}$ is equal to\begin{array}{1 1}(A)\;\sin(b-a)\log\begin{vmatrix}\frac{\sin(x-b)}{\sin(x-a)}\end{vmatrix}+C & (B)\;cosec(b-a)\log\begin{vmatrix}\frac{\sin(x-a)}{\sin(x-b)}\end{vmatrix}+C \\(C)\;cosec(b-a)\log\begin{vmatrix}\frac{\sin(x-b)}{\sin(x-a)}\end{vmatrix}+C &(D)\;\sin(b-a)\log\begin{vmatrix}\frac{\sin(x-a)}{\sin(x-b)}\end{vmatrix}+C \end{array}

Can you answer this question?

Toolbox:
• $\sin(A+B)=\sin A\cos B+\cos A\sin B$
• $\sin(A-B)=\sin A\cos B-\cos A\sin B$
• $\int \cot xdx=\log\mid\sin x\mid+c.$
Step 1:
Let $I=\int\large\frac{dx}{\sin (x-a)\sin(x-b)}$
Multiply and divide by $\sin(a-b)$
$\quad=\int\large\frac{\sin(a-b)}{\sin(a-b)\sin(x-a)\sin(x-b)}\normalsize dx.$
$\sin(a-b)$ in the numerator can be written as $\sin[(x-b)-(x-a)]$
$\quad=\large\frac{1}{\sin(a-b)}\int\large\frac{\sin(x-b)-(x-a)}{\sin(x-a)\sin(x-b)}\normalsize dx.$
$\sin(A-B)=\sin A\cos B-\cos A\sin B$
$\quad=\large\frac{1}{\sin(a-b)}\int\large\frac{\sin(x-b)\cos(x-a)-\cos(x-b)\sin(x-a)}{\sin(x-a)\sin(x-b)}\normalsize dx.$
$\large\frac{\cos(x-a)}{\sin(x-a)}=\normalsize\cot(x-a)$ and $\large\frac{\cos(x-b)}{\sin(x-b)}=\normalsize \cot(x-b)$
$\quad=\large\frac{1}{\sin(a-b)}$$\int[\cot(x-a)-\cot(x-b)] dx. Step 2: Now on integrating we get \quad=\large\frac{1}{\sin(a-b)}$$[\log\mid\sin(x-a)\mid-\log\mid\sin(x-b)\mid]+c .$
But $\large\frac{1}{\sin(a-b)}$$=cosec(a-b)$
$\;\;=cosec(a-b)\log\begin{vmatrix}\large\frac{\sin(x-a)}{\sin(x-b)}\end{vmatrix}+c.$
$\;\;=cosec(b-a)\log\begin{vmatrix}\large\frac{\sin(x-b)}{\sin(x-a)}\end{vmatrix}+c.$
Hence the correct option is C.
answered Apr 24, 2013