$$\int \frac{1}{3\sin(x)+2\cos(x)+3}\ \text{d}x$$ 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?
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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!
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 $\tan\frac{x}{2}=t\implies \frac{1}{2}\sec^2\frac{x}{2}\ dx=dt$ $$=\int \frac{2\ dt}{t^2-4}$$ $$=\frac{2}{4}\ln\left|\frac{t-2}{t+2}\right|+C$$ $$=\color{blue}{\frac{1}{2}\ln\left|\frac{\tan\frac{x}{2}-2}{\tan\frac{x}{2}+2}\right|+C}$$