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Please help me evaluate: $$ \int\frac{dx}{\sin(x+a)\sin(x+b)} $$

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2 Answers 2

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Hint: Multiply and divide by $\sin(b-a)$.

Further Hint:

Write $\sin(b-a) = \sin((x-a)-(x-b))$

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The given integral is: $$\int\frac{dx}{\sin(x+a)\sin(x+b)}$$

The given integral can write:

$$\int\frac{dx}{\sin(x+a)\sin(x+b)}=\int\frac{\sin(x+a)}{\sin(x+b)}\cdot\frac{dx}{\sin^2(x+a)}$$

We substition $$\frac{\sin(x+a)}{\sin(x+b)}=t$$

By the substition of the above have:

$$\frac{dx}{\sin^2(x+a)}=\frac{dt}{\sin(a-b)}$$

Now have:

$$\int\frac{dx}{\sin(x+a)\sin(x+b)}=\int\frac{\sin(x+a)}{\sin(x+b)}\cdot\frac{dx}{\sin^2(x+a)}=\frac{1}{\sin(a-b)}\int\frac{dt}{t}=\frac{1}{\sin(a-b)}\ln |t|=\frac{1}{\sin(a-b)}\ln|\frac{\sin(x+a)}{\sin(x+b)}|+C$$

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    $\begingroup$ This is supposed to be a homework question, so I presume you should leave the proposer with some room to do it himself... $\endgroup$ Commented Sep 4, 2012 at 20:14