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Integrate the following :

$$\int \frac{\sin x}{\sin4x}dx$$

My approach :

$$=\int \frac{\sin x}{2\sin2x \cos2x}dx$$

$$= \int \frac{\sin x}{4\sin x \cos x \cos2x}dx$$ $$= \int \frac{1}{4\cos x \cos2x}dx $$ [ didn't get the hint here ] then I did the following way.

$$= \int \frac{\sin x}{4\sin x \cos x (\cos2x) }dx$$

Now putting $\cos x =t$ $\Rightarrow -\sin x \, dx = dt $

$$= \int \frac{-dt}{4\sqrt{1-t^2} t(2t^2-1)}$$ [Using : $ \cos2x = 2\cos^2x-1]$

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  • $\begingroup$ In the second last step set $\sin x = t$, then the integral becomes $\int \frac{dt}{(1-t^2)(1-2t^2)}$ $\endgroup$ – r9m Mar 17 '14 at 3:07
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From the step before the last step you had it as:

$$\eqalign{\int\frac{dx}{\cos x\cos2x} &= \int\frac{dx}{\cos x(2(\cos x)^2 - 1))}\cr & = \int\frac{1}{\cos x} - \frac{2\cos x}{(2(\cos x)^2 - 1)}dx\cr & = \int\frac{1}{\cos x} - \frac{2\cos x}{1 - 2(\sin x)^2}dx\ .\cr}$$

From now you can integrate 1/cosx = secx and its antiderivative is well known.

The second term: let u = sinx, then du = cosxdx, and use fration decomposition to continue.

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    $\begingroup$ I think there is a problem of negative sign in step II of your solution viz. $\frac{1}{cosx}-\frac{2cosx}{2cos^2x-1} \neq \frac{1}{cosx(2cos^2x-1)}$ ... is it so please confirm. $\endgroup$ – sultan Mar 17 '14 at 4:26
  • $\begingroup$ Please correct your answer. The mistake has been pointed out by @Sultan $\endgroup$ – Archer Aug 3 '18 at 8:14
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$$\frac1{\cos x\cos2x}=\frac{\cos x}{\cos^2x(\cos2x)}=\frac{\cos x}{(1-\sin^2x)(1-2\sin^2x)} $$

Setting $\displaystyle\sin x=u$ in $\displaystyle I=\int\frac{dx}{\cos x\cos2x},$

$\displaystyle I=\int\frac{du}{(1-u^2)(1-2u^2)} =\frac12\int\frac{du}{(1-u^2)\left(\dfrac12-u^2\right)} $

$\displaystyle I=\frac22\int\frac{(1-u^2)-\left(\dfrac12-u^2\right)}{(1-u^2)\left(\dfrac12-u^2\right)}du=\int\frac{du}{1-u^2}-\int\frac{du}{\dfrac12-u^2}$

Can you take it home from here?

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  • $\begingroup$ Why has the above wrong answer received so many upvotes? $\endgroup$ – Archer Aug 3 '18 at 6:49
  • $\begingroup$ @Abcd, Please pinpoint the mistake $\endgroup$ – lab bhattacharjee Aug 3 '18 at 7:12
  • $\begingroup$ I am referring to DeepSea's answer. Isn't that wrong? $\endgroup$ – Archer Aug 3 '18 at 7:30
  • $\begingroup$ @Abcd, The comment should have been posted in that answer. Yes, incorrect & one can rectify it $\endgroup$ – lab bhattacharjee Aug 3 '18 at 7:34

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