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Find $\displaystyle \int \dfrac{1}{2\sin(x)-3\cos(x)}dx$.

My book said to solve this by saying $u = \tan \left(\dfrac{x}{2} \right)$ since $\cos(x) = \dfrac{1-u^2}{1+u^2}$ and $\sin(x) = \dfrac{2u}{1+u^2}$. I don't see how this will help since $du = \dfrac{1}{\cos(x)+1}dx $. How will we get that in the integrand?

Puzzled417
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4 Answers4

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Hint: (alternative way)

You can write the expression $2\sin(x)-3\cos(x)$ as $\sqrt{13}\sin\left(x-\tan^{-1}\left(\tfrac32\right)\right)$. You'll thus have to integrate: $$\dfrac{1}{\sqrt{13}}\csc\left(x-\tan^{-1}\left(\tfrac32\right)\right).$$

Community
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Workaholic
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  • I do not understand hoe it makes it simpler – Shailesh Dec 18 '15 at 23:54
  • @Shailesh It transformed it into something of the form $A\csc(x+B)$, one can easily show that thus its integral is of the form $$ A\left{\ln\left(\sin\dfrac{B+x}{2}\right) -\ln\left(\cos\dfrac{B+x}{2}\right)\right}+\tt C.$$ – Workaholic Dec 19 '15 at 20:34
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    Thanks. I did not see that the tan inverse term was a constant, – Shailesh Dec 19 '15 at 23:28
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Notice, $$\int \frac{1}{2\sin x-3\cos x}\ dx$$ $$=\int \frac{1}{2\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}-3\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}\ dx$$ $$=\int \frac{1+\tan^2\frac{x}{2}}{3\tan^2\frac{x}{2}+4\tan\frac{x}{2}-3}\ dx$$ $$=\frac13\int \frac{\sec^2\frac{x}{2}}{\tan^2\frac{x}{2}+\frac{4}{3}\tan\frac{x}{2}-1}\ dx$$ $$=\frac23\int \frac{d\left(\tan\frac{x}{2}\right)}{\left(\tan\frac{x}{2}+\frac23\right)^2-\left(\frac{\sqrt{13}}{3}\right)^2}\ dx$$ $$=\frac{2}{3\times 2\frac{\sqrt {13}}{3}}\ln\left|\frac{\tan\frac{x}{2}+\frac23-\frac{\sqrt{13}}{3}}{\tan\frac{x}{2}+\frac23+\frac{\sqrt{13}}{3}}\right|+C$$

$$=\frac{1}{\sqrt {13}}\ln\left|\frac{3\tan\frac{x}{2}+2-\sqrt{13}}{3\tan\frac{x}{2}+2+\sqrt{13}}\right|+C$$

Charles Brillon
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Harish Chandra Rajpoot
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2

Hint:

the suggested substitution works well because: $$ \cos x+1=\frac{1-u^2}{1+u^2}+1=\frac{2}{1+u^2} $$ so: $$ du=\frac{1}{2\cos^2(x/2)}dx= \frac{dx}{\cos x+1} \Rightarrow dx=\frac{2\,du}{1+u^2} $$ the integral becomes: $$ 2\int\frac{du}{3u^2+4u-3} $$ that, completing the square, becomes: $$ 2\int\frac{du}{(\sqrt{3}u+2\sqrt{3}/3)^2-13/3} $$ that can be solved with the substitution $t=\sqrt{3}u+2\sqrt{3}/3$ and factorizing $13/3$.

Michael Hardy
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Emilio Novati
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The substitution works because you get $$ x=2\arctan u $$ so $$ dx=\frac{2}{1+u^2}\,du $$ Since $$ 2\sin x-3\cos x=2\frac{2u}{1+u^2}-3\frac{1-u^2}{1+u^2}= \frac{3u^2- 4u -3}{1+u^2} $$ the integral becomes $$ \int\frac{1+u^2}{3u^2- 4u -3}\frac{2}{1+u^2}\,du= 2\int\frac{1}{3u^2-4u -3}\,du $$ that's solvable by partial fractions.

A different way is to write $$ 2\sin x-3\cos x=A\sin(x-\varphi) $$ that becomes $$ 2\sin x-3\cos x=A\sin x\cos\varphi-A\cos x\sin\varphi $$ so we can set $$ A\cos\varphi=2,\quad A\sin\varphi=3 $$ and so $A^2=13$. There is a unique angle $\varphi\in[0,2\pi)$ such that $$ \sin\varphi=\frac{3}{\sqrt{13}},\quad \cos\varphi=\frac{2}{\sqrt{13}} $$ Now the integral becomes $$ \sqrt{13}\int\frac{1}{\sin(x-\varphi)}\,dx $$ and we can do the substitution $x-\varphi=2u$, so $dx=2\,du$ and the integral is \begin{align} \sqrt{13}\int\frac{1}{\sin u\cos u}\,du &=\sqrt{13}\int\frac{\sin^2u+\cos^2u}{\sin u\cos u}\,du\\ &=\sqrt{13}\left(\int\frac{\sin u}{\cos u}\,du +\int\frac{\cos u}{\sin u}\,du\right)\\ &=\sqrt{13}(\log|\cos u|-\log|\sin u|)+c \end{align} and it's just tedious to do the back substitution.

egreg
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