Trigonometric integration 3.

Question

$\int \frac{\sqrt{3cos2x-1}}{cosx}$

ATTEMPT:

I did the following substitution

Let $sinx=t$

$cosxdx=dt$

$I=\frac{\sqrt{3(1-2t^2)-1}}{1-t^2}=\frac{\sqrt{2-6t^2}}{1-t^2}$

Now let $t=\frac{1}{z}$

$dt=-\frac{1}{z^2}dz$

Now $I=\color{red}{-} \frac{\sqrt{2z^2-6}}{z(z^2-1)}$

Lastly substitute $2z^2-6=p^2$

Finally $I=\color{red}{-}2\int \frac{p^2}{(p^2+6)(p^2+4)}dp$ which can be easily be done by using Partial fractions.

Substituting back from $p$ to $z$ to $t$ and finally to $sinx.$

I got $I=2arctan\sqrt{\frac{2-6sin^2x}{4sin^2x}}\color{red}{-}\sqrt{6}arctan\sqrt{\frac{2-6sin^2x}{6sin^2x}}$

But the answer given in the text is $\sqrt{6}arcsin(\sqrt{3}sinx)-2arctan(\frac{2sinx}{\sqrt{3Cos2x-1}})$

Where did i go wrong?

Answer 1

I think you are using a wrong substitution.

Try $Cos2x=1-2Sin^2x$. Then do the substitution using $t=Cosx$.

Answer 2

Not sure I can follow your last substitution, but I think the mistake was in dropping a sign when you changed variables from $t$ to $z$. It appears your answer is off by a sign.

Draw a right triangle and label one degree $\alpha$.Make the side opposite $\alpha$ a length of $\sqrt3\sin x$ and the hypoteneuse a length of $1$. Then $\sin\alpha=\sqrt3\sin x$ and $\alpha=\sin^{-1}(\sqrt3\sin x)$. By the Pythagorean theorem, the third side has length $\sqrt{1-3\sin^2x}.$ Note that $\tan\alpha=\frac{\sqrt3\sin x}{\sqrt{1-3\sin^2x}}=\sqrt\frac{3\sin^2x}{1-3\sin^2x}$. The other acute angle, however has a tangent equal to $\sqrt\frac{1-3\sin^2x}{3\sin^2x}.$ Therefore, $\tan^{-1}\sqrt\frac{1-3\sin^2x}{3\sin^2x}=\frac\pi2-\alpha$. Similarly, you can find that

$$\tan^{-1}\sqrt\frac{2-6\sin^2x}{4\sin^2x}=\tan^{-1}\frac{\sqrt{3\cos2x-1}}{2\sin x}=\frac\pi2-\tan^{-1}\frac{2\sin x}{\sqrt{3\cos2x-1}}$$

Therefore,

$$\sqrt6\tan^{-1}\sqrt\frac{1-3\sin^2x}{3\sin^2 x}-2\tan^{-1}\sqrt\frac{1-3\sin^2x}{2\sin^2x}=$$

$$\sqrt6(\frac\pi2-\sin^{-1}(\sqrt3\sin x))$$

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