1
$\begingroup$

$$\int \frac{1}{3\sin(x)+2\cos(x)+3}\ \text{d}x$$ This is one of the question which appeared in my exam today and I tried solving this but just couldn't solve it. I tried converting sine and cosine in terms of $\tan\left(\frac{x}{2}\right)$ but didn't help very well..

Does anyone have idea how to evaluate this integral?

$\endgroup$
  • 1
    $\begingroup$ use the tan half angle formulas $\endgroup$ – Dr. Sonnhard Graubner Oct 20 '15 at 9:32
  • 1
    $\begingroup$ Please explain how the $u=\tan(x/2)$ "didn't help very well". It should. $\endgroup$ – mickep Oct 20 '15 at 9:33
  • 2
    $\begingroup$ after two minutes you know that this will not help? $\endgroup$ – Dr. Sonnhard Graubner Oct 20 '15 at 9:37
  • 2
    $\begingroup$ see here en.wikibooks.org/wiki/Calculus/Integration_techniques/… $\endgroup$ – Dr. Sonnhard Graubner Oct 20 '15 at 9:37
  • $\begingroup$ @Dr.SonnhardGraubner. I know that we use this type of substitution for this form of integral, but isn't it too lengthy? I got the answer, but there has to be a shorter method also? Please can you point me towards it? $\endgroup$ – Aditya Agarwal Oct 20 '15 at 9:55
1
$\begingroup$

Notice, $$\int \frac{1}{3\sin x+2\cos x+3}\ dx$$ $$=\int \frac{1}{3\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}+2\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}+3}\ dx$$ $$=\int \frac{1+\tan^2\frac{x}{2}}{\tan^2\frac{x}{2}+6\tan\frac{x}{2}+5}\ dx$$ $$=\int \frac{\sec^2\frac{x}{2}}{\left(\tan\frac{x}{2}+3\right)^2-4}\ dx$$ Let $\tan\frac{x}{2}=t\implies \frac{1}{2}\sec^2\frac{x}{2}\ dx=dt$ $$=\int \frac{2\ dt}{t^2-4}$$ $$=\frac{2}{4}\ln\left|\frac{t-2}{t+2}\right|+C$$ $$=\color{blue}{\frac{1}{2}\ln\left|\frac{\tan\frac{x}{2}-2}{\tan\frac{x}{2}+2}\right|+C}$$

$\endgroup$
1
$\begingroup$

HINT:

$$\int \frac{1}{3\sin(x)+2\cos(x)+3}\text{d}x =$$


Substitute $u=\tan\left(\frac{x}{2}\right)$ and $\text{d}u=\frac{1}{2}\sec^2\left(\frac{x}{2}\right)\text{d}x$. Than transform the integrand using the substitutions $\sin(x)=\frac{2u}{u^2+1}$, $\cos(x)=\frac{1-u^2}{u^2+1}$ and $\text{d}x=\frac{2}{u^2+1}\text{d}u$:


$$\int \frac{2}{(u^2+1)\cdot\left(\frac{6u}{u^2+1}+\frac{2(1-u^2)}{u^2+1}+3\right)}\text{d}u =$$ $$\int \frac{2}{u^2+6u+5}\text{d}u =$$ $$2\int \frac{1}{(u+3)^2-4}\text{d}u=$$


Substitute $s=u+3$ and $\text{d}s=\text{d}u$:


$$2\int \frac{1}{s^2-4}\text{d}s=$$ $$2\int -\frac{1}{4\left(1-\frac{s^2}{4}\right)}\text{d}s=$$ $$-\frac{1}{2}\int \frac{1}{1-\frac{s^2}{4}}\text{d}s=$$


Substitute $p=\frac{s}{2}$ and $\text{d}p=\frac{1}{2}\text{d}s$:


$$-\int \frac{1}{1-p^2}\text{d}p=$$ $$-\tanh^{-1}(p)+C$$

Substitute everything back and you'll get the final answer!

$\endgroup$
  • 1
    $\begingroup$ Sure you've to substitute everything back $\endgroup$ – Jan Oct 20 '15 at 9:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.