Question

Find the integral of$\frac{(\sin x+\cos x)}{[3+\sin 2x]}dx$.

Answer 1

Find the integral of the given function

Given integral:$\int \frac{(\sin x+\cos x)}{[3+\sin 2x]}dx$

Integral of $\frac{(\sin x+\cos x)}{[3+\sin 2x]}dx$ can be written as:

$$\begin{gathered} \int \frac{(\sin x+\cos x)}{[3+\sin 2x]}dx=\int \frac{(\sin x+\cos x)}{\left[\right(4-1)+\sin 2x]}dx \\ =\int \frac{(\sin x+\cos x)}{[4-(1-\sin 2x]}dx \\ =\int \frac{(\sin x+\cos x)}{[4-({\sin }^{2}x+{\cos }^{2}x-\sin 2x]}dx\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{x}\mathbf{+}\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{2}}\mathbf{x}\mathbf{=}\mathbf{1}\mathbf{]} \\ =\int \frac{(\sin x+\cos x)}{[4-({\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x]}dx \\ =\int \frac{(sinx+cosx)}{[4-{(sinx-cosx)}^2]}dx....\left(1\right)\left[\because {(a–b)}^2=a^2–2ab+b^2\right] \end{gathered}$$

Let,

$\sin x–\cos x=t....\left(2\right)$

On differentiating both the sides, we get

$(\cos x+\sin x)dx=dt....\left(3\right)$

On substituting equation (2) and equation (3) in equation (1) we get,

$\int \frac{(\sin x+\cos x)}{[3+\sin 2x]}dx=\int \frac{dt}{(2^2–t^2)}$

We know that,

$\int \frac{1}{a^2-x^2}dx=\frac{1}{2a}\log \left|\frac{a+x}{a-x}\right|+C$, where $C$ is the integration constant.

So,

$\int \frac{dt}{(2^2–t^2)}$$=\frac{1}{2\left(2\right)}\log \left|\frac{2+t}{2-t}\right|+C$, where $C$ is the integration constant.

$=\frac{1}{4}\log \left|\frac{2+\sin x-\mathrm{co}sx}{2-\sin x+\mathrm{co}sx}\right|+C$

$\therefore \int \frac{(\sin x+\cos x)}{[3+\sin 2x]}dx$$=\frac{1}{4}\log \left|\frac{2+\sin x-\mathrm{co}sx}{2-\sin x+\mathrm{co}sx}\right|+C$

Hence, the integral of $\frac{(\sin x+\cos x)}{[3+\sin 2x]}dx$ is $\frac{1}{4}\log \left|\frac{2+\sin x-\mathrm{co}sx}{2-\sin x+\mathrm{co}sx}\right|$, where $C$ is the integration constant.